
Construct a pendulum that is 1 meter long and measure its period for small oscillations using a phone clock.
Vary the oscillation angle and plot the period as a function of oscillation angle.
Measure the period for small oscillations for a pendulum with lengths of .5, 1, and 2 meters.
P_{½} = Period for a length of .5 meters P_{1} = Period for a length of 1 meters P_{2} = Period for a length of 2 metersUsing the measured values, calculate P_{1}/P_{½}, P_{2}/P_{1}, and P_{2}/P_{½}.
The analytic result for the period for small oscillations is:
Pendulum length = L = 1 meter Gravity = g = 9.8 m/s^{2} Period = T = 2 π (L/g)^{1/2} = 2.006 secondsThe oscillation period increases as the angle increases.
Era Method for measuring speed Renaissance Use a pendulum clock to measure time and a ruler to measure distance 20th century Use a pocket watch or phone clock to measure time and a ruler to measure distance 21st century Film the object and analyze the video framebyframeRoll a ball across a table and measure its speed using the stopwatch and the phone video methods. What would you estimate is the error for each method?
Velocity = V Time = T Position = X = V TBy viewing a video framebyframe you can measure the position and time of the ball for a set of different times. For example,
Frame Time Position (s) (m) 0 .0 .10 12 .5 .21 24 1.0 .32 36 1.5 .43 48 2.0 .54 60 2.5 .65 72 3.0 .76 Frame rate = 24 frames/secondThe velocity at Time=.75 can be approximated as:
Time of first measurement = T_{1} = .5 Time of second measurement = T_{2} = 1.0 Position at first measurement = X_{1} = .21 Position at second measurement = X_{2} = .32 Time difference = T = T_{2}  T_{1} = 1.0  .5 = .5 Position difference = X = X_{2}  X_{1} = .32  .21 = .11 Velocity at Time=.75 = V = X / T = .22 meters/second
Roll two balls toward each other so that they collide headon and rebound in the opposite direction, and use a phone video to measure the quantities listed below.
Blue ball: Initially on the left and moving to the right Red ball: Initially on the right and moving to the left Momentum = Mass * Velocity Kinetic energy = ½ * Mass * Velocity^{2} Mass of blue ball = M_{1} Mass of red ball = M_{2} Initial velocity of blue ball = V_{1i} Initial velocity of red ball = v_{2i} Final velocity of blue ball = V_{1f} Final velocity of red ball = v_{2f} Initial momentum of blue ball = Q_{1i} = M_{1} V_{1i} Initial momentum of red ball = Q_{2i} = M_{2} V_{2i} Final momentum of blue ball = Q_{1f} = M_{1} V_{1f} Final momentum of red ball = Q_{2f} = M_{2} V_{2f} Initial energy of blue ball = E_{1i} = ½ M_{1} V_{1i}^{2} Initial energy of red ball = E_{2i} = ½ M_{2} V_{2i}^{2} Final energy of blue ball = E_{1f} = ½ M_{1} V_{1f}^{2} Final energy of red ball = E_{2f} = ½ M_{2} V_{2f}^{2} Total initial momentum = Q_{i} = Q_{1i} + Q_{2i} Total final momentum = Q_{f} = Q_{1f} + Q_{2f} Total initial energy = E_{i} = E_{1i} + E_{2i} Total final energy = E_{f} = E_{1f} + E_{2f} Energy ratio = E_{r} = E_{f} / E_{i}Momentum is conserved in collisions: Initial momentum = final momentum.
Collisions usually convert some energy to heat: Final energy < Initial energy
Gravity energy = Mass * g * Height Kinetic energy = ½ * Mass * Velocity^{2}Drop a ball from rest and measure the height of the first bounce.
Ball mass = M Gravity constant = g = 9.8 meters/second^{2} Initial height = X_{i} Final height = X_{f} (Maximum height after the first dim Initial gravity energy = E_{i} = M g X_{i} Final gravity energy = E_{f} = M g X_{f} Height ratio = X_{r} = X_{f} / X_{i} Energy ratio = E_{r} = E_{f} / E_{i}Plot the height ratio (X_{r}) as a function of height (X_{i}).
Suppose you climb a set of stairs.
Height = Height of a set of stairs Mass = Mass of a person Gravity = 9.8 m/s^{2} Energy = Mass * Gravity * Height Time = Time required to climb the stairs Power = Energy / Time = Mass * Gravity * Height / Time = Mass * Gravity * Vertical velocity Agility = Power / MassClimb 3 flights of stairs and measure the above quantities.
If a 100 kg person eats 3000 Calories in one day then
Energy = 3000 Calories * 4.2e3 Joules/Calorie = 12.6 MJoules Power = Energy / Time = 12.6e6 Joules / 1 Day = 12.6e6 Joules / 86400 seconds = 146 Watts or Joules/second Agility = Power / Mass = 1.46 Watts/kg
Measure your maximum running speed and calculate your kinetic energy.
Mass = Mass of a person Velocity = Maximum running velocity Kinetic energy = ½ Mass * Velocity^{2}
g = 9.8 meters/second^{2} Gravity energy = Mass * g * Height Kinetic energy = ½ Mass * Velocity^{2}How much gravitational potential energy does it take to raise an object vertically from the surface of the Earth to a height of 400 km (the height of the space station)?
The space station orbits at 7.8 km/s. How much kinetic energy does 1 kilogram of matter have if it is moving at this speed?
Using Wikipedia, how much energy does one kilogram of gasoline have?
If an object starts from rest at X=0 and undergoes constant acceleration then after time T,
Time = T Acceleration = A Velocity = V = A T Position = X = .5 A T^{2} = V^{2} / (2 A) Velocity = Change in position / Change in time Acceleration = Change in velocity / Change in time
Record a video of a ball dropping and measure the height (X) and time (T) to reach the floor. Calculate the gravitational acceleraton. Use X=.5 meters and 2.0 meters.
A = 2 X / T^{2}
Roll a sphere down an inclined plane and measure the distance traveled for the first 4 seconds. Let
X_{1} = Distance traveled after 1 second X_{2} = Distance traveled after 2 seconds X_{3} = Distance traveled after 3 seconds X_{4} = Distance traveled after 4 secondsIf the acceleration is constant then
R_{2} = X_{2}/X_{1} = 4 R_{3} = X_{3}/X_{1} = 9 R_{4} = X_{4}/X_{1} = 16Measure X1, X2, X3, X4, and calculate R2, R3, R4.
Construct a pendulum using as large a length and mass as possible. The Earth's rotaton causes the pendulum to precess like the animation above, although the precession is exaggerated in the animation.
Start the pendulum and observe its direction anle and then observe the direction angle one hour later.
Q = Rate of change of the direction angle of a pendulum = 360 * sin(Latitude) degrees/day = 360 * sin(40.667 degrees) degrees/day For New York City = 234.6 degrees/day = 9.78 degrees/hour New York City Latitude = 40.667 degrees North New York City Longitude = 73.933 degrees West
100 Zhang Heng constructs a seismometer using pendulums that was capable of detecting the direction of an Earthquake. 1500 Pendulums are used for power for machines such as saws, bellows, and pumps. 1582 Galileo finds that the period of a pendulum is independent of mass and oscillation angle, if the angle is small. 1636 Mersenne and Descartes find that a pendulum is not quite isochronous. Its period increased somewhat with its amplitude. 1656 Huygens builds the first pendulum clock, with a precision of 15 seconds per day. Previous devices had a precision of 15 minutes per day. 1658 Huygens publishes the result that pendulum rods expand when heated. This was the principal error in pendulum clocks. 1670 Previous to 1670 the verge escapement was used, which requires a large angle. The anchor escapement mechanism is developed in 1670, which allows for a smaller angle. This increased the precision because the oscillation period is independent of angle for small angles. 1673 Huygens publishes a treatise on pendulums. 1721 Methods are developed for compensating for thermal expansion error. 1726 Gridiron pendulum developed, improving precision to 1 second per day. 1851 Foucault shows that a pendulum can be used to measure the rotation period of the Earth. The penulum swings in a fixed frame and the Earth rotates with respect to this frame. In the Earth frame the pendulum appears to precess. 1921 Quartz electronic oscillator developed 1927 First quartz clocks developed, which were more precise than pendulum clocks.
Construct a balance scale using any materials that would have been available to Newton.
Collect a set of identical coins to use as standard masses. Dimes are ideal because they have the smallest mass.
Measure the mass of one of the balls from the list below in units of coin masses and then use the table of coins to convert it to kg. What is the relative error?
Suppose there are N coins on the left side of the balance and N+1 coins on the right, with all coins being identical. If N is small then the scale can tell the difference and if N is large it can't. What is the largest value of N for which you can tell the difference between N coins and N+1 coins?
We can define a "resolution" for the scale as 1/N. For example, if a scale has a maximum mass of 1 kg and it can resolve down to 1 gram, then its resolution is .001 kg / 1 kg = 0.001.
For a nickel, measure the mass, diameter, and thickness, and calculate the volume and density. Compare it to the table below.
Volume = Thickness * π * (Diameter / 2)^{2} Density = Mass / Volume
Mass Diameter Thickness Density Copper Nickel Zinc Manganese (g) (mm) (mm) Dime 2.268 17.91 1.35 8.85 .9167 .0833 Penny 2.5 19.05 1.52 7.23 .025 .975 Copper plated Nickel 5.000 21.21 1.95 7.89 .75 .25 Quarter 5.670 24.26 1.75 9.72 .9167 .0833 1/2 dollar 11.340 30.61 2.15 10.20 .9167 .0833 Dollar 8.100 26.5 2.00 9.73 .885 .02 .06 .035 Plated with manganese brass Dollar bill 1.0 155.956 .11 .088 Height = 66.294 mm
In ancient times, gold was an ideal currency because it was hard to counterfeit. No other element known had a density that was nearly as large.
Silver can be counterfeited because lead is more dense and cheaper.
The metals known to ancient civilizations were:
Density Known to ancient (g/cm^{3}) civilizations Zinc 7.1 * Manganese 7.2 Tin 7.3 * Iron 7.9 * Nickel 8.9 Copper 9.0 * Bismuth 9.8 * Silver 10.5 * Lead 11.3 * Mercury 13.5 * Tungsten 19.2 Gold 19.3 * Platinum 21.4 Osmium 22.6 Densest elementIn ancient times, gold could be countereited to a limited degree because mass and volume could not be measured precisely. One could shave a small amount of gold from a coin, small enough so that the change in volume is undetectable.
Once tungsten was discovered in 1783 it became easy to counterfeit gold.
Newton was Master of the Mint and he placed the United Kingdom on the gold standard. He was the Sherlock Holmes of his era and he caught all the counterfeiters.
In this figure, ball sizes are in scale with each other and court sizes are in scale with each other. Ball sizes are magnified by 10 with respect to court sizes.
The distance from the back of the court to the ball is the characteristic distance the ball travels before losing half its speed to air drag.
Ball Ball Court Court Ball diameter Mass length width density (mm) (g) (m) (m) (g/cm^{3}) Ping pong 40 2.7 2.74 1.525 .081 Squash 40 24 9.75 6.4 .716 Golf 43 46 1.10 Badminton 54 5.1 13.4 5.18 .062 Racquetball 57 40 12.22 6.10 .413 Billiards 59 163 2.84 1.42 1.52 Tennis 67 58 23.77 8.23 .368 Baseball 74.5 146 .675 Pitcherbatter distance = 19.4 m Whiffle 76 45 .196 Football 178 420 91.44 48.76 .142 Rugby 191 435 100 70 .119 Bowling 217 7260 18.29 1.05 1.36 Soccer 220 432 105 68 .078 Basketball 239 624 28 15 .087 Cannonball 220 14000 7.9 For an iron cannonball
Dot size = Atomic radius = (AtomicMass / Density)^{1/3}For gases, the density at boiling point is used.
Earliest Shear Melt Density known use Strength (K) (g/cm^{3}) (year) (GPa) Wood < 10000 15  .9 Rock < 10000 Carbon < 10000 Diamond < 10000 534 3800 3.5 Gold < 10000 27 1337 19.3 Silver < 10000 30 1235 10.5 Sulfur < 10000 Copper 9000 48 1358 9.0 Lead 6400 6 601 11.3 Brass 5000 ~40 Copper + Zinc Bronze 3500 ~40 Copper + Tin Tin 3000 18 505 7.3 Antimony 3000 20 904 6.7 Mercury 2000 0 234 13.5 Iron 1200 82 1811 7.9 Arsenic 1649 8 1090 5.7 Cobalt 1735 75 1768 8.9 First metal discovered since iron Platinum 1735 61 2041 21.4 Zinc 1746 43 693 7.2 Tungsten 1783 161 3695 19.2 Chromium 1798 115 2180 7.2 Stone age Antiquity Copper age 9000 Bronze age 3500 Iron age 1200 Carbon age 1987 Jimmy Connors switches from a steel to a graphite racketBronze holds an edge better than copper and it is more corrosion resistant.
Horizontal axis: Density Vertical axis: Shear modulus / Density (Strengthtoweight ratio)Strengthtoweight ratio is important for swords. Iron makes a better sword than copper or bronze.
Horizontal axis: Density Vertical axis: Shear moduus / Density (Strengthtoweight ratio)Beryllium is beyond the top of the plot.
Metals with a strengthtoweight ratio less than lead are not included, except for mercury.
$\theta $ = Angle in radians (dimensionless) X = Arc distance around the circle in meters (the red line in the figure) R = Radius of the circle in meters X = $\theta $ R
π is defined as the ratio of the circumference to the diameter.
Full circle = 360 degrees = 2 $\pi $ radians
1 radian = 57.3 degrees
1 degree = .0175 radians
Angle in degrees = (180/π) * Angle in radians
Angle in radians = (π/180) * Angle in degrees
Radius = R Angle = $\theta $ (radians) X coordinate = X = R cos($\theta $) Y coordinate = Y = R sin($\theta $)
Let (X,Y) be a point on a circle of radius R.
θ = Angle of the point (X,Y) in radians X = R cos(θ) Y = R sin(θ) Y/X = tan(θ)If θ is close to zero then
X ~ R Y << X Y << R sin(θ) ~ θ tan(θ) ~ θThe "small angle approximation" is
Y/X ~ θ
A person with 20/20 vision can distinguish parallel lines that are spaced by an angle of .0003 radians, about 3 times the diffraction limit. Text can be resolved down to an angle of .0015 radians.
Resolution Resolution Diopters for parallel for letters (meters^{1}) lines (radians) (radians) 20/20 .0003 .0015 0 20/40 .0006 .0030 1 20/80 .0012 .0060 2 20/150 .0022 .011 3 20/300 .0045 .025 4 20/400 .0060 .030 5 20/500 .0075 .038 6"Diopters" is a measure of the lens required to correct vision to 20/20.
The closest distance your eyes can comfortably focus is 20 cm. If a computer screen is at this distance then the minimum resolvable pixel size is
Pixel size = Angle * Distance = .0003 * .2 = .00006 meters = .06 mmFor a screen that is 10 cm tall this corresponds to 1670 pixels. This is referred to as a "retinal display".
Measure your visual resolution angle for the following situations:
Resolving pairs of dots
Resolving parallel lines
Resolving Letters (both for dim and bright light)
Resolving pixels on a phone
All waves diffract, including sound and light. Light passing through your pupil is diffracted and this sets the limit of the resolution of the eye.
Diameter of a human pupil = D = .005 meters
Wavelength of green light = W = 5.5*10$7$ meters
Characteristic diffraction angle = θ = .00013 radians = 1.22 * W/D (for a circular aperture)
The colossal squid is up to 14 meters long, has eyes up to 27 cm in diameter, and inhabits the ocean at depths of up to 2 km.
There are two ways to measuring parallax: "without background" and "with background". The presence of a background improves the precision that is possible.
Without background:
With background:
Place two observer marks on the floor around 1 meter apart and place a target mark on the other side of the room, at least 8 meters away from the observer marks. Arrange the observer marks to be perpendicular to the target mark, like in the figure above.
X = Distance between the observer marks D = Distance from an observer mark to the target mark (should be the same for both observer marks)
Align the flat end of a protractor with the line between the observer marks, and measure the angles from the observer marks to the target mark. Both angles should be near 90 degrees.
θ_{1} = Angle from observer mark #1 to the target mark θ_{2} = Angle from observer mark #2 to the target mark θ = θ_{2}  θ_{1}Using the small angle approximation,
θ = X / Dwhere θ is in radians. Measure X and θ and calculate D with the smallangle approximation. Also measure D.
Look out the lab window and find two buildings that overlap each other. The far building should be much further away than the near building. Use Google Maps to find the distances to the buildings.
The near building is the target for which we will measure the distance, and the far building is the background that allows us to measure precise angles.
Select two vantage points from inside the lab that are as far apart as possible and that can both see the buildings, and measure the distance between them. Measure the difference in the angle that the two vantages perceive of the near building, and calculate the distance to the near building.
Distance to near building = Distance between the vantage points / Difference in angle
Eratosthenes produced a measurement of the Earth that was accurate to 2 percent.
Ptolemy developed a system of latitude and longitude for mapping the world. His map covered 1/4 of the globe and was the standard until the Renaissance.
Find a long pole and use it to measure the angle of the sun with respect to due south. Take the measurement at the moment when the sun is highest in the sky. Use a pendulum bob to ensure that the pole is precisely vertical. At the same time, have an accomplice at a different latitude perform the same measurement. Use Google maps to determine the distance between you and your accomplice in the NorthSouth direction, and use the measurements to calculate the radius of the Earth.
The radius of the Earth is
θ_{1} = Angle of the shadow measured in New York City in degrees θ_{2} = Angle of the shadow measured by the accomplice X = Distance between you and your accomplice in the latitude direction = EarthRadius * θ_{1}θ_{2} π / 180 (meters) New York City Latitude = 40.667 degrees North New York City Longitude = 73.933 degrees West Earth radius = 6371 km
Montreal and Manhattan have nearly the same longitude, which means that Montreal is directly north of Manhattan.
Manhattan latitude = 40.667 degrees North Manhattan longitude = 73.933 degrees West Earth radius = R = 6371 km Montreal latitude = 45.500 degrees North Montreal longitude = 73.567 degrees West MontrealManhattan latitude difference = θ MontrealManhattan distance = X = R θWhat is the difference in latitude between Montreal and Manhattan in radians?
In 1714, the British Parliament established the "Longitude Prize" for anyone who could find an accurate method for determing longitude at sea.
John Harrison solved the problem by developing precise clocks but Parliament refused to pay out. In 1772, Harrison gave one of his clocks to King George III who personally tested it and found it to be accurate to 1/3 of one second per day. King George III advised Harrison to petition Parliament for the full prize after threatening to appear in person to dress them down.
Maskelyne was the chairman of the board responsible for awarding the Longitude prize and he refused to award it to Harrison. Maskelyne developed the "Lunar distance method" for determing longitude, which was decisively defeated by Harrison's clocks in a test at Barbados. Also, James Cook abandoned the lunar distance method after his first world voyage and used Harrison's clocks for his 2nd and 3rd voyages.
From Wikipedia: "Cook's log is full of praise for the watch and the charts of the southern Pacific Ocean he made with its use were remarkably accurate."
Maskelyne held the post of "Astronomer Royal" and was hence in charge of awarding the Longitude Prize. He opposed awarding it to Harrison and Harrison was instead paid for his chronometers by an act of parliament.
Measure the time of sunset and also have an accomplice at a different longitude do the same measurement. Use the measurements to calculate the difference in longitude and use Google maps to find the exact value.
T_{1} = Time that you measure for sunset in hours T_{2} = Time that your accomplice measures for sunset in hours L_{1} = Your longitude in degrees L_{2} = Your accomplice's longitude in degrees 15 * (T_{1}  T_{2}) = L_{1}  L_{2}
Build a scale model of the sun, Mercury, Venus, the Earth, and the Earth's moon, with sizes and distances to scale. Choose a length of at least 50 meters for the distance from the Earth to the sun. Use Wikipedia for numbers.
Construct a scale model of the following systems:
The Earth, the moon, and the L2 Lagrange point.
The Milky Way, the Large Magellanic Cloud, Andromeda, and M87 (galaxy at the center of the Virgo Cluster).
A violin, a viola, a cello, a bass, a guitar, and a bass guitar. Only size matters here, not distance.
Suppose you have a set of measurements
Trial Measurement number result 1 1.232 2 1.251 3 1.256 4 1.245 5 1.233 6 1.238 7 1.433The numbers cluster around the value "1.24" except for measurement #7 "1.433", which is an "outlier". Generally the outliers are removed and the error is computed from the wellbaheaved numbers. Usually the outliers are errors, although on occasion it can turn out that the outlier is the correct measurement and the seemingly wellbahaved numbers are in error. There is no general rule for this. One has to be careful. In the following calculations we exclude the outlier.
Suppose we have N measurements X_{j}. The mean is
Mean = N^{1} * ∑_{j} X_{j} = (1/6) * (1.232 + 1.251 + 1.256 + 1.245 + 1.233 + 1.238) = 1.242The "Gaussian error" is
Error^{2} = N^{1} * ∑_{j} (X_{j}  Mean)^{2} = 6^{1} * [ (1.2321.242)^{2} + (1.2511.242)^{2} + (1.2561.242)^{2} + (1.2451.242)^{2} + (1.2331.242)^{2} + (1.2381.242)^{2} ] = .0090If we were to include the outlier then it would dominate the calculation, rendering the other measurements meaningless.
The measurement is quoted as
Measured value = Mean + Gaussian error = 1.242 + .0090
Suppose the length of an object is measured several times, with the results in meters being:
X_{1} = 2.553 X_{2} = 2.534 X_{3} = 2.536 X_{4} = 2.563 X_{5} = 2.541 X_{6} = 2.544 X_{7} = 2.560 X_{8} = 2.539What is the mean and the Gaussian error? Plot the data to show how it is distributed.
Energy = E (Joules) Mass = M (kg) Volume = Vol (m^{3}) Time = T (seconds) Time required for the battery to drain Power = P = E / T (Watts) Power delivered by the battery Energy/Volume = E_{vol} = E / Vol Energy/Mass = E_{mass}= E / MBattery energies are often quoted in WattHours or AmpHours.
Voltage = V = 3.7 Volts for a Lithium battery Current = I (Current supplied by the battery in Amps) Power = P = I V (Power delivered by the battery in Watts) 1 WattHour = Energy associated with a power of 1 Watt for a duration of 1 hour = Power * Time = 1 Watt * 3600 Seconds = 1 Joule/second * 3600 seconds = 3600 Joules 1 AmpHour = Energy associated with a current of 1 Amps for a duration of 1 hour = Power * Time = Current * Voltage * Time = 1 Amp * 3.7 Volts * 3600 Seconds = 13320 JoulesFor example,
20 WattHours = 20 Watts * 3600 seconds = 72000 Joules 5.4 AmpHours = 5.4 * 13320 Joules = 72000 Joules
For a phone or tablet battery, use the printed value for
WattHours or AmpHours to calculate the energy.
Measure the mass and volume and calculate the energy/mass and energy/volume.
Data for batteries from Amazon.com.
Energy Energy Length Width Height Energy Energy $ Energy/$ density (MJ) (m) (m) (m) (Wh) (Ah) (kJ/$) (MJ/m^{3}) Anker Astro E3 900 .137 136.9 67.3 16.5 10 2.7 22 6.2 Poweradd Pilot Pro 680 .426 185.4 121.9 27.9 118.4 32 130 3.3 Ravpower 23000 650 .306 185 124.5 20.3 85.1 23 100 3.1 1 kJ = 10^{3} Joules 1 MJ = 10^{6} Joules
Force can be measured using mass and gravity.
Mass of an object = M Gravity acceleration at the Earth's surface = g = 9.8 meters/second^{2} Gravity force at the Earth's surface = F = M gA 1 kilogram object in Earth gravity exerts a force of 9.8 Newtons, which is 2.205 pounds.
X = Length of a string under zero force x = Change in string length when a force is applied X+x = Total length of the string when a force is applied K = Spring constant Force = Force on the spring = K x (Hooke's law)Using any string or rope available, construct a plot of Force as a function of x, all the way up to the breaking point. Set the string length "X" equal to 1 meter if possible.
In the region of low x, what is the value of K?
The elasticity of a wire depends on its intrinsic stiffness and on its cross sectional area.
The tensile modulus characterizes the stiffness of a wire and it is proportional to the spring constant.
For a wire,
X = Length of wire under zero tension force x = Increase in length of the wire when a tension force is applied K = Spring constant Force = Tension force on the wire = K x Area = Crosssectional area of the wire Pressure= Force / Area (Pressure, measured in Pascals or Newtons/meter^{2}) Strain = Fractional change in length of the wire (dimensionless) = x/X Modulus = Tensile modulus or "Young's modulus" for the wire material (Pascals) = Pressure / StrainStarting from Hooke's law, we can derive an equation relating the modulus to the spring constant.
Force = Pressure * Area = K * x = K * X * x / X = K * X * Strain = Modulus * Area * Strain Pressure = (K * X / Area) * Strain = Modulus * Strain Modulus = K X / Area K = Modulus * Area / X
Choose a wire made out of any material, such as fishing line, a strip of duct tape, or a shoelace. Hang the wire from the tower and add weights to the wire until it breaks. Meaure and calculate the following:
Length of the wire under zero weight = X Length change of the wire at breaking point = x Change in length required to break wire Cross sectional area of the wire = A Hanging mass required to break the wire = M Gravity constant = g = 9.8 m/s^{2} Force required to break the wire = F = M g Spring constant = K = F / x Tensile stiffness = P_{stiff} = F X / (x A) (Pascals) Tensile flexibility = T_{flex} = x / X Tensile strength = P_{strong}= F / A (Pascals) Energy per volume of the wire material = e = ½ P_{stiff} T^{2}_{flex} (Joules/meter^{3}) 1 Pascal = 1 Newton/meter^{2} = 1 Joule/meter^{3}
Tensile Breaking Breaking Tough Tough/ Brinell Density modulus pressure strain density (GPa) (g/cm^{3}) (GPa) (GPa) (MPa) (J/kg) Beryllium 287 .448 .0016 .350 189 .6 1.85 Magnesium 45 .232 .0052 .598 344 .26 1.74 Aluminum 70 .050 .00071 .018 15 .245 2.70 Titanium 120 .37 .0031 .570 54 .72 4.51 Copper 130 .210 .0016 .170 19 .87 8.96 Bronze 120 .800 .0067 2.667 300 8.9 Iron 211 .35 .0017 .290 37 .49 7.87 Steel 250 .55 .0022 .605 77 7.9 Stainless 250 .86 .0034 1.479 185 8.0 Chromium 279 .282 .00101 .143 199 1.12 7.15 Molybdenum 330 .324 .00098 .159 15 1.5 10.28 Silver 83 .170 .0020 .174 17 .024 10.49 Tungsten 441 1.51 .0037 2.585 134 2.57 19.25 Osmium 590 1.00 .0018 .893 40 3.92 22.59 Gold 78 .127 .0016 .103 5.3 .24 19.30 Lead 16 .012 .00075 .045 3.8 .44 11.34 Rubber .1 .016 Nylon 3 .075 .025 .938 815 1.15 Carbon fiber 181 1.600 .0088 7.07 4040 1.75 Kevlar 100 3.76 Zylon 180 5.80 1.56 Nanorope ~1000 3.6 .0036 6.5 4980 1.3 Graphene 1050 160 .152 12190 12190000 1.0 Glass 45 .033 2.53 Concrete 30 .005 2.7 Granite 70 .025 2.7 Marble 70 .015 2.6 Bone 14 .130 .0093 604 377 1.6 Ironwood 21 .181 .0086 780 650 1.2 Human hair .380 Spider silk 1.0 1.3 Sapphire 345 1.9 .0055 5232 1315 3.98 Diamond 1220 2.8 .0023 3210 920 1200 3.5 Toughness = Energy / Volume Toughness / Density = Energy / MassA climbing rope should have a large toughness/density. It should absorb a lot of energy and it should be light enough to carry.
If a force is applied to the center of a beam then it bends into a circular shape. The tensile modulus and tensile strength can be measured by measuring the deflection.
Measure the following:
Length of the beam = X (largest dimension of the beam) Width of the beam = Y Height of the beam = Z (parallel to the force applied) Force required to break the beam = F (at center of beam and in the direction of the Z axis) Beam deflection when it breaks = x (displacement of the center of the beam) Spring constant = K = F / x Tensile modulus = Y = (3/16) F X^{3} / (X Y x Z^{3}) = (3/16) K X^{3} / (X Y Z^{3}) Internal strain when it breaks = S = 4 Z x / X^{2} Tensile strength = P = S Y (internal pressure when it breaks) Energy/Volume when it breaks = e = .5 * Y S^{2} Mass of the beam = M Density of the beam = D = M / (X Y Z) Energy/Mass = e / D
Gravity constant = g = 9.8 meters/second^{2} Mass of a sled resting on a table = M_{sled} Mass of a weight hanging from the wire = M_{hang} Force of the sled on the table = F_{sled} = M_{sled} g Force on the weight hanging from the wire = F_{hang} = M_{hang} g Area of the sled in contact with the table = A Minimum sideways force to move the sled = Q F_{hang} Coefficient of friction of the sled = Q = F_{hang} / F_{sled} = M_{hang} / M_{sled}Construct a sled and place masses on the sled. Attach a wire to the sled and use the wire to generate a sideways force. Add weights on the wire until the sled moves and measure the required weight. Calculate the friction coefficient.
Plot the coefficient of friction as a function of sled mass.
Using fixed sled mass, plot the friction coefficient as a function of sled area.
The friction coefficient depends on the types of surfaces used.
Surface Surface Friction #1 #2 coefficient Concrete Rubber 1.0 Steel Steel .8 Wood Wood .4 Metal Wood .3 Concrete Rubber (wet) .3 Wood Ice .05 Ice Ice .05 Steel Ice .03Try experiments with different kinds of surfaces and measure the coefficient of friction.
A pulley allows one to change the direction of a force.
This lab uses the My Solar System simulaton at phet.colorado.edu.
Set up a simulation with the following parameters.
Mass Position Velocity X Y X Y Body 1 100. 0 0 0 0 Star Body 2 1. 100 0 0 V Planet V_{c} = Velocity for which the planet orbits as a circle. V_{e} = Escape velocity. Minimum velocity to escape.Try varying V and using trial and error, estimate the vales of V_{c} and V_{e}. What does the formula below predict?
If the planet velocity is changed from the Y direction to the X direction, what is V_{e}?
If the planet's X position is changed to 50 then what is V_{c}?
R = Planet X coordinate V_{c} = Velocity for a circular orbit V_{e} = Velocity for escape G = Gravity constant = 10000 for the simulator A = Gravitational acceleration M = Star mass m = Planet massFor a planet on a circular orbit,
Gravitational Force = Centripetal force G M m / R^{2} = m V_{c}^{2} / R V_{c} = (GM/R)^{1/2}For a planet to escape the star,
Gravitational energy = Kinetic energy G M m / R = .5 m V_{e}^{2} V_{e} = √2 * V_{c} = (2GM/R)^{1/2}
If two planets are too close together then they will interfere gravitationally.
Using the simulator, set up a system with 2 planets.
Mass Position Velocity X Y X Y Body 1 100. 0 0 0 0 Star Body 2 .01 100 0 0 100 Planet 1 Body 3 .01 x 0 0 v Planet 2To give Planet 2 a circular orbit, use
v = 1000 / √x x v 100 100 105 98 110 95 115 93 120 91 125 89 130 88 135 86 140 85 145 83 150 82If "x" is close to 100 then the planets interfere gravitationally, and if "x" is far from 100 the planets ignore each other.
Run the simulation for values of x ranging from 100 to 150 and determine the minimum value of x for the planets to not interfere.
You can travel between planets with a "Hohmann maneuver". You start from the inner circular orbit, fire the rocket, cruise on an elliptical "transfer orbit" to the outer orbit, and then fire the rocket again to put the rocket into the outer circular orbit.
The Earth and Mars system can be simulated using the following values. Both the Earth and Mars are on circular orbits.
Mass Position Velocity X Y X Y Body 1 100. 0 0 0 0 Sun Body 2 .000219 100 0 0 100 Earth Body 3 .000032 152 0 0 81 MarsIn a Hohmann maneuver a spaceship starts at the Earth and fires its rockets in the Y direction, in the same direction as the Earth's velocity.
V_{earth} = Earth velocity V_{launch} = Departure velocity of the rocket with respect to the Earth V_{total} = Total rocket velocity = V_{earth} + V_{launch}If V_{launch} has the right value then the rocket's orbit will graze Mars' orbit, and this represents the minimum amount of fuel.
If V_{launch} is too low then the rocket won't make it to Mars.
If V_{launch} is too high then the rocket overshoots Mars' orbit. This gets you to Mars faster but uses more fuel than the grazing orbit.
In the simulation, increase the Earth's "Y" velocity (V_{total}) until you find the value that causes the Earth to graze Mars' orbit. What is this velocity?
A planet "Tatooine" can be added halfway between Venus and Earth with
Mass Position Velocity X Y X Y Body 1 100. 0 0 0 0 Sun Body 2 .000219 72 0 0 118 Venus Body 3 .000303 86 0 0 108 Tatooine, a clone of the Earth that is closer to the sun Body 4 .000303 100 0 0 100 EarthIs this system stable? How large do you have to make the mass of the middle planet to make the system unstable?
Using the Lunar lander simulation, try to land the spacecraft using a minimum of fuel. What is the minimum fuel needed for a soft landing? Describe the strategy you used.
In the Android app "Osmos" you can experiment with maneuvering a spaceship in a gravitational potential. Once the app is started go to level 3 "solar".
The game is like Saturn's ring. You are a snowball in the ring surrounded by other snowballs and you can observe the differential motion between nearby snowballs. You can also change your velocity and observe the effect on your orbit.
If you are on a circular orbit of radius R and you want to change to a circular orbit of radius 2R, what is the most efficient strategy? How would you draw a diagram to illustrate this?
The game is also like a model of an accretion disk. In the sun's accretion disk, objects accumulated by gravity into planets and the same thing happens in Osmos. Large objects tend to accumulate faster than small objects and the end result is a set of planets with widelyseparated orbits. This phenomenon is mirrored in Osmos because in the game, large objects tend to accumulate faster than small objects.
Suppose you play the game with the purpose of observing how accretion works. Move the spaceship to an orbit in the Kuiper belt so that it doesn't interfere with the accretion. After the accretion has finished, what does the result look like?
Film a ball rolling alongside a meter stick and analyze the video framebyframe to evaluate time and position. For example,
Time Position (s) (m) .0 .000 .5 .100 1.0 .195 1.5 .285 2.0 .370 2.5 .450 3.0 .525The velocity at Time=.25 can be approximated as
Velocity = Change in position / Change in time = (.100  .000) / (.5  .0) = .2 meters/secondThe velocity at Time=.75 can be approximated as
Velocity = (.195  .100) / (1.0  .5) = .19 meters/secondContinuing, we can generate a table of velocities.
Time Position Velocity (s) (m) (m/s) .0 .000 .25 .20 .5 .100 .75 .19 1.0 .195 1.25 .18 1.5 .285 1.75 .17 2.0 .370 2.25 .16 2.5 .450 2.75 .15 3.0 .525From the table you can tell that the object starts out with a velocity of .20 and decelerates.
The acceleration at Time=.50 can be approximated as:
Acceleration = Change in velocity / Change in time = (.19  .20) / (.75  .25) = .02 meters/second^{2}We can continue the procedure to produce a table of velocities and accelerations.
Time Position Velocity Acceleration (s) (m) (m/s) (m/s^{2}) .0 .000 .25 .2 .5 .100 .02 .75 .19 1.0 .195 .02 1.25 .18 1.5 .285 .02 1.75 .17 2.0 .370 .02 2.25 .16 2.5 .450 .02 2.75 .15 3.0 .525
Make a video of a ball rolling across a table and use the above procedure to generate a table of positions, velocities, and accelerations.
Plot the following:
Position as a function of time
Velocity as a function of time
Acceleration as a function of time
The drag force for an object moving through air is
Object mass = M Object area = A Crosssectional area Object velocity = V Air density = d = 1.22 kg/m^{3} Drag constant = C Dimensionless and usually equal to 1 Drag force = F = .5 C d A V^{2}
For a falling balloon,
Gravitational acceleration = g = 9.8 m/s^{2} Gravitational force on the balloon = F_{grav} = M g Air density = d = 1.22 kg/m^{3} Balloon crosssectional area = A Balloon velocity = V Balloon drag force = F_{drag} = ½ C d A V^{2} Balloon drag coefficient = C = F_{drag} / (½ d A V^{2}) Balloon terminal velocity = V_{term} = (2 M g / C / d / A)^{2}If a balloon is falling at terminal velocity then the gravitational force is equal to the drag force.
F_{grav} = F_{drag} M g = ½ C d A V_{term}^{2}Drop a balloon and measure its mass, terminal velocity, and crosssectional area. Use the formula to calculate the drag coefficient.
Add mass to the balloon so that its new mass is 4 times the old mass, and measure the new terminal velocity. What is Q?
Q = (Terminal velocity for mass "4M") / (Terminal velocity for mass "M")
Suppose you want to estimate how far a soccer ball travels before air drag slows it down. For a soccer ball,
Ball mass = M = .437 kg Ball radius = R = .110 meters Ball area = A = .0380 meters^{2} = π R^{2} Ball density = D = 78.4 kg/meter^{3} Air density = d = 1.22 kg/meter^{3} Newton length = L = 9.6 meters = M/d/A Characteristic distance the ball travels before slowing down Air mass = m = A L d Air mass that the ball passes through after distance LNewton observed that the characteristic distance L is such that
m = MHence
L = M / (d A) = (4/3) R D / dThe depth of the penalty box is 16.45 meters (18 yards). Any shot taken outside the penalty box slows down before reaching the goal.
Newton was also the first to observe the "Magnus effect", where spin causes a ball to curve.
Type of light Wavelength (nm) Threshold for cell damage 300 Magenta limit of vision 400 Magenta 440 Blue 480 Cyan 520 Green 555 Yellow 620 Red limit of photosynthesis 680 Red 700 Red limit of vision 750
Humans can see light from 400 nm to 750 nm.
Light is harmful if it has a wavelength smaller than 300 nm.
Photosynthesis can use light from 300 nm to 680 nm, except for the green light at 555 nm.
The Blackbody radiation simualtion at phet.colorado.edu plots the blackbody spectrum as a function of temperature. The area under the curve is the amount of energy produced by the blackbody. You can subdivide the energy into bands. For example,
Energy Largest Smallest type wavelength wavelength (nm) (nm) Infrared Infinity 750 Visual 750 400 UV 400 0 Total energy Infinity 0
You can use the simulator to estimate the energy of each type by estimating the area under the curve for the appropriate wavelength range.
In the figure above,
UV energy = Area of the gray area to the left Visual energy = Area of the rainbow zone Infrared energy = Area of the gray area to the rightThe sun has a temperature of 6000 Kelvin. Using the simulator, estimate the values of
Infrared energy / Total energy Visual energy / Total energy UV energy / Total energyThe estimate doesn't have to be overly precise. An eyeball estimate will do.
Estimate the temperature of a blackbody for which
UV energy / Total energy = 1/100
Build a bridge using the following materials:
Wood (tongue depressor, toothpick, chopstick, etc.)
Paper (regular paper or file folder paper)
Superglue
Cotton string
Duct tape
Plastic straw
To test the bridge, two tables will be placed 30 cm apart and the bridge will be placed across the gap. Masses will be loaded on the bridge until it breaks, and the score is the breaking is given as follows.
M_{break} = Mass required to break the bridge M_{bridge} = Mass of the bridge (40 grams maximum) S = Score of the bridge = M_{break} / M_{bridge}
Build a tower 30 cm high. Weights will be placed on the tower until the tower collapses and the score will be calculated similarly as the bridge score.
M_{break} = Mass required to break the tower M_{tower} = Mass of the tower (40 grams maximum) S = Score of the tower = M_{break} / M_{tower}
Build a catapult (trebuchet) to launch a projectile. You can design the catapult so that it launches the projectile when a string is cut.
A catapult consists of a beam to support the masses, and a tower to support the beam. The beam should be as long as possible and the tower should be as high as possible, and both should be lightweight so that they can be carried by horses by a medieval army.
M_{cat} = Mass of the catapult (40 grams maximum) M_{drive}= Mass of the object used to drive the catapult (can have any value) M_{proj} = Mass of the projectile launched by the catapult (can have any value) X = Distance the projectile travels, measured from the front of the catapult S = Score of the catapult = X M_{proj}
The drive mass is typically much larger than the projectile mass.
The properties of a wave are
Frequency = F (seconds^{1}) Wavelength = W (meters) Wavespeed = V (meters/second) Period = T (seconds) = The time it takes for one wavelength to pass byWave equations:
F W = V F T = 1
Length of a train car = W = 10 meters (The wavelength) Speed of the train = V = 20 meters/second (The wavespeed) Frequency = F = 2 Hertz (Number of train cars passing by per second) Period = T = .5 seconds (the time it takes for one train car to pass by)
Speed of sound at sea level = V = 340 meters/second Frequency of a violin A string = F = 440 Hertz Wavelength of a sound wave = W = .77 meters = V/F Wave period = T = .0023 seconds
A wave on a string moves at constant speed and reflects at the boundaries.
For a violin Astring,
Frequency = F = 440 Hertz Length = L = .32 meters Time for one round trip of the wave = T = .0023 s = 2 L / V = 1/F Speed of the wave on the string = V = 688 m/s = F / (2L) String equation: 2 L V = F
In a reed instrument, a puff of air enters the pipe, which closes the
reed because of the Bernoulli effect. A pressure pulse travels to the other
and and back and when it returns it opens the reed, allowing another puff of
air to enter the pipe and repeat the cycle.
Green dots indicate the frequencies of open strings.
An orchestral bass and a bass guitar have the same string tunings.
The range of organs is variable and typically extends beyond the piano in both the high and low direction.
Violins, violas, and cellos are tuned in fifths. String basses, guitars, and bass guitars are tuned in fourths. Pianos are tuned with equal tuning.
Frequency (Hertz) Violin E 660 = 440 * (3/2) Violin A 440 Violin D 293 = 440 / (3/2) Violin G 196 = 440 / (3/2)^{2} Viola A 440 Same as a violin A Viola D 293 Viola G 196 Viola C 130 Cello A 220 One octave below a viola A Cello D 147 Cello G 98 Cello C 65 String bass G 98 = 55 * (3/2)^{2} String bass D 73 = 55 * (3/2) String bass A 55 3 octaves below a violin A String bass E 41 = 55 / (3/2) Guitar E 330 Guitar B 244 Guitar G 196 = 110 * (4/3)^{2} Guitar D 147 = 110 * (4/3) Guitar A 110 = 2 octaves below a violin A of 440 Hertz Guitar E 82.5 = 110 / (4/3)When an orchestra tunes, the concertmaster plays an A and then everyone tunes their A strings. Then the other strings are tuned in fifths starting from the A.
A bass guitar is tuned like a string bass.
According to legend Bach used a supersized viola, the "Viola Pomposa"
If two notes are played at the same time then we hear the sum of the waveforms.
If two notes are played such that the frequency of the high note is twice that of the low note then this is an octave. The wavelength of the high note is half that of the low note.
Color Frequency Wavelength Orange 220 Hertz 1 Red 440 Hertz 1/2Because the red and orange waves match up after a distance of 1 the blue note is periodic. This makes it easy for your ear to process.
If we double both frequencies then it also sounds like an octave. The shape of the blue wave is preserved.
Color Frequency Wavelength Orange 440 Hertz 1/2 Red 880 Hertz 1/4When two simultaneous pitches are played our ear is sensitive to the frequency ratio. For both of the above octaves the ratio of the high frequency to the low frequency is 2.
440 / 220 = 2 880 / 440 = 2If we are talking about frequency ratios and not absolute frequencies then for simplicity we can set the lower frequency to 1.
Frequency Normalized frequency 220 1 440 2 880 4
The octave, fifth, fourth, major third, and minor third are all
periodic and sound harmonious.
The tritone is not periodic and sounds dissonant.
If two notes in an interval have frequencies such that
Frequency of top note / Frequency of bottom note = I / J where I and J are small integersthen the combined note is periodic. The smaller the integers I and J, the more noticeable the periodicity and the more harmonious the interval. This is why fifths and fourths sound more resonant than thirds.
Fret Note Interval Equal Just Major Minor tuning tuning scale scale 0 A Unison 1.000 1.000 = 1/1 * * Harmonious 1 Bb Minor second 1.059 Dissonant 2 B Major second 1.122 1.125 = 9/8 * * Dissonant 3 C Minor third 1.189 1.200 = 6/5 * Weakly harmonious 4 C# Major third 1.260 1.250 = 5/4 * Weakly harmonious 5 D Fourth 1.335 1.333 = 4/3 * * Harmonious 6 Eb Tritone 1.414 Dissonant 7 E Fifth 1.498 1.500 = 3/2 * * Harmonious 8 F Minor sixth 1.587 1.600 = 8/5 * Weakly harmonious 9 F# Sixth 1.682 1.667 = 5/3 * Weakly harmonious 10 G Minor seventh 1.782 * Dissonant 11 Ab Major seventh 1.888 * Dissonant 12 A Octave 2.000 2.000 = 2/1 * * HarmoniousThe red dots correspond to the locations of the frets on a ukelele, which is tuned with "equal tuning". Guitars also use equal tuning. A violin and a ukelele have the same string length and string tuning.
The green dots correspond to "just tuning", which is used by fretless instruments. Just tuning is based on integer frequency ratios.
If the note "A" is played together with the notes of the 12tone scale then the result is
The major and minor scales favor the harmonious notes.
In equal tuning, the frequency ratio of an interval is
Frequency ratio = 2^{Fret/12} where "Fret" is an integerEqual tuning is based on equal frequency ratios. Just tuning adjusts the frequencies to correspond to the nearest integer ratio. For example, in equal tuning, the frequency ratio of a fifth is 1.498 and just tuning changes it to 1.500 = 3/2.
The notes [Bb, B, Eb, G, Ab] cannot be expressed as a ratio of small integers and so they sound dissonant when played together with an A.
For the 12 tone scale, equal tuning and just tuning are nearly identical.
Frequency Wavelength (Hertz) (meters) 20 15 Lower limit of human frequency sensitivity 41 8.3 Lowestfrequency string on a string bass or bass guitar 65 2.52 Lowestfrequency string on a cello 131 2.52 Lowestfrequency string on a viola 440 .75 The Astring on a violin 660 .75 The Estring on a violin (highestfrequency string) 20000 .016 Upper limit of human hearing
A ukelele Astring has a frequency of 440 Hertz and the frets are set according to the following table.
Note Fret Frequency A 0 440 Open Astring .1 442.5 Largest frequency that sounds indistinguishable from 440 Hertz Bb 1 466 Half step B 2 494 Whole step C 3 523 C# 4 554 D 5 587 Perfect fourth Eb 6 622 Tritone E 7 659 Perfect fifth F 8 698 F# 9 740 G 10 784 G# 11 831 A 12 880 OctaveFret ".1" is an imaginary fret that represents the highest note that is indistinguishable to the ear from the open string. It is located about 1/10 of the way to the first half step.
If a string has a frequency of 880 Hertz then frets are such that:
Note Fret Frequency (Hertz) A 0 880 .1 885 Largest frequency that sounds indistinguishable from 880 Hertz Bb 1 932 Half step B 2 988 Whole step ...
If a wave is linear then it propagates without distortion.
If a wave is linear then waves add linearly and oppositelytraveling waves pass through each other without distortion.
If two waves are added they can interfere constructively or destructively, depending on the phase between them.
If a speaker system has 2 speakers you can sense the interference by moving around the room. There will be loud spots and quiet spots.
The more speakers, the less noticeable the interference.
Noisecancelling headphones use the speakers to generate sound that cancels incoming sound.
Two waves traveling in opposite directions create a standing wave.
Waves on a string simulation at phet.colorado.edu
Whan a wave on a string encounters an endpoint it reflects with the waveform
preserved and the amplitude reversed.
When an string is played it creates a set of standing waves.
Length of a string = L Speed of a wave on the string = V Overtone number = N An integer in the set {1, 2, 3, 4, ...} Wavelength of an overtone = W = 2 L / N Frequency of an overtone = F = V / W = V N / (2L) N = 1 corresponds to the fundamental tone N = 2 is one octave above the fundamental N = 3 is one octave plus one fifth above the fundamental N = 4 is two octaves above the fundamentalAudio: overtones
For example, the overtones of an Astring with a frequency of 440 Hertz are
Overtone Frequency Note 1 440 A 2 880 A 3 1320 E 4 1760 A 5 2200 C# 6 2640 E 7 3080 G 8 3520 AOvertone simulation at phet.colorado.edu
A spectrum tells you the power that is present in each overtone.
The first row is the waveform, the second row is the waveform expanded in time, and the third row is the spectrum. The spectrum reveals the frequencies of the overtones. In the panel on the lower left the frequencies are 300, 600, 900, 1200, etc. In the panel on the lower right there are no overtones.
A quality instrument is rich in overtones.
A waveform can be represented as an amplitude as a function of time or as an amplitude as a function of frequency. A "Fourier transform" allows you to go back and forth between these representations. A "spectrum" tells you how much power is present at each frequency.
Fourier transform simulation at phet.colorado.edu
Software such as "Garage Band", and the Android app "FrequenSee" can record music and display the spectrum.
Every instrument produces sound with a different character. The sound can be characterized either with the waveform or with the spectrum
In the following plots the white curve is the waveform and the orange dots are the spectrum.
Obtain a spectrum app for your phone. "FrequenSee" works for Android and "Garage Band" works for iPhone. Find any 1D resonator (such as a pitchfork, a string, or a bottle) and strike it so that it rings. Use the spectrum app to measure the resonant frequencies. The resonant frequencies will appear as spikes in the spectrum. Measure as many spikes as you can.
F_{1} = Frequency of the lowestfrequanty spike F_{2} = Frequency of the spike with the next highest frequency after F_{1} F_{3} = Frequency of the spike with the next highest frequency after F_{2} F_{4} = etc. R_{2} = F_{2} / F_{1} R_{3} = F_{3} / F_{1} R_{4} = F_{4} / F_{1}Calculate R_{2}, R_{3}, R_{4}, etc., for as many spikes as the resonator has.
Try the experiment with different kinds of resonators. Use any resonator you can find.
1D resonators: Strings, rods, and bottles.
2D resonators: Drums, plates, the body of a stringed instrument.
3D resonators: Interior of a soccer ball or globe.
A wave on a string moves at constant speed and reflects at the boundaries.
For a violin Astring,
Frequency of the lowestfrequency overtone = F = = 440 Hertz (= F_{1} from above) Length of the string = L = = .32 meters Time for one round trip of the wave = T = 2 L / V = 1/F = .0023 s Speed of the wave on the string = V = F / (2L) = 688 m/s String equation: 2 L V = F_{}For each of the 1D resonators from the previous lab, measure the length of the resonator and the frequency of the lowestfrequency note and use them to calculate the wavespeed V.
Build a musical instrument using rubber bands for strings. Invent a mechanism for tuning the strings, such as like the pegs on a violin.
Give the instrument two identical strings and tune them to have the same frequency. Measure the length and frequency of the string and calculate the wavespeed. The frequency should be at least 200 Hertz to produce a useable tone.
Suppose you play the left string open and the right string with a finger down.
L_{1} = Length of the open left string L_{2} = Length of the right string, from one end to the finger This is the active part of the string that can vibrate when you pluck it. L_{1} > L_{2} R = Frequency ratio between the two notes. = L_{1} / L_{2}Pythagoras tried different values of R and found that some values sound harmonious and others sound dissonant.
Try all possible values of R from 1 to 4 and look for harmonious values. Record any values you find.
If you have an actual stringed instrument, try the experiment with the instrument. If you have a wind or brass instrument then try playing it together with another instrument.
An electrical pickup device will be provided (costs $2 at Radio Shack) that can deliver the sound to a speaker, which will help in hearing the tone. This allows lower frequencies to be used.
Build a musical instrument, either acoustic or electric (electrical pickups will be provided). If it is acoustic, try to make the instrument as loud as possible, especially for low notes (it's a challange to make low notes loud). If it is electric, try finding novel resonators and record the sound.
Conduct an experiment to measure the sensitivity of human frequency perception. For example, suppose you use a sound generator to produce a frequency of 440 Hertz and then slowly change the frequency until you notice that the frequency has changed.
Original frequency = F = 440 Hertz Frequency sensitivity = F_{rez} Resolution for measuring a frequency of "F" Frequency sensitivity ratio = R = F_{rez} / F Relative frequency resolution that is independent of FOnline tone generator
Suppose you play a note with a frequency of "F" and slowly raise it to a higher frequency "f".
If fF < F_{rez} then "f" sounds the same as "F" If fF > F_{rez} then "f" sounds different from "F"Conduct an experiment to measure the value of R for a range of frequencies F = 440 and 880 Hertz.
Let θ be the characteristic angle for which you can sense the direction of a sound.
For the experiment there is noisemaker and a listener. The noisemaker makes a sound while the listener has his eyes closed. The listener points to the direction he thinks the sound is coming from, then opens his eyes and measures the angle error. Do 6 trials to produce 6 numbers and then put these numbers through the error lab procedure to calculate the Gaussian error.
Obtain an Online tone generator.
Using a smartphone power spectrum app such as FrequenSee (Android) or Garage Band (Apple), play a note at 220 Hertz and draw the power spectrum for various speakers, such as:
Headphones
Smartphone
Tablet
Laptop
Desktop
The large speaker in the lab
Using any speaker, start from a frequency of 440 Hertz and observe the peak of the lowestfrequency overtone. Decrease the frequency and watch the peak. At the moment it vanishes, record the frequency "F_{bass}".
F_{bass} = Lowest frequency that a speaker can produce D = Diameter of the speakerMeasure F_{bass} and D for each of the speakers listed above.
The walls of an anechoic chamber absorb all sound.
The absorbers are pointy to minimize the reflection of sound.
The information rate for sound is kilobytes/second and the rate for vision is megabytes/second.
Build an anechoic chamber to be as silent as possible and measure the decibel level. What measures did you have to take to reduce noise?
Obtain an app for measuring sound intensity and perform measurements in any place you might be in Manhattan. Record the results. Is there any place other than Central Park where you can't hear cars?
Use the app to measure the decibel reduction in sound when it passes through a wall. Play a sound in an adjacent room and measure the sound level in the adjecent room and the lab room.
Use a sound intensity app to measure the loudness of various instruments. Place the microphone a standard 1 meter from the instrument for each instrument. Measure the intensity of the lowest note and each octave above it.
After a string is plucked the amplitude of the oscillations decreases with time. The larger the damping the faster the amplitude decays.
F = Frequency of the string T = Time for one oscillation of the string = 1/F T_{damp}= Characteristic timescale for vibrations to damp q = "Quality" parameter of the string = Characteristic number of oscillations required for the string to damp = T_{damp} / T = T_{damp} * FThe smaller the damping the larger the value of q. For most musical instruments, q > 100.
For various resonators, measure T_{damp} and F and use them to estimate
the quality factor q = T_{damp} F.
For example, you can strike a resonator and estimate how long it rings before damping out, or you can record the waveform with Garage Band and use it to estimate T_{damp}.
You can break a wine glass by singing at the same frequency as the glass's resonanant frequency. An expensive wineglass has a large quality factor. The larger the quality factor, the easier it is to break the glass by singing.
Get an account on Wikipedia and improve a page. Pages in need of improvement include:
Greenhouses: Water and fertilizer requirements for crops.
Water quality of rivers, expressed as "Biological oxygen demand".
Sewage treatment: costs, fertilizer yield, biomass yield.
Irrigation: Data on water requirements with and without drip irrigation.
Seawater greenhouses: Data from existing greenhouses.
Desalination: Data from existing plants.
Emergency management: Disaster risk and monetary losses. Cost of prevention.
Iron fertilization of the oceans.
Urban forestry: Data for tree growth rates, trunk size, and height.
Solar cells: prices, efficiencies, and element requirements.
Wind turbines: prices, efficiencies, and element requirements.
Electric power distribution.
Prefabricated homes: Data for sizes, prices, and raw materials.
Patents: cost of solar cells, wind turbines, and smartphones.
Any topic relating to the presidential election.
Any topic from the history of science.
"The Hum"