AP Physics 1 Unit 7: Oscillations

3. (a) 1 point. This is really three separate investigations. In each case,

students should measure the period repeatedly and then take the

mean value.

1 point for procedures similar to these:

- Amplitude versus period. Take a period measurement for 5 to 10

different amplitudes while keeping the mass and length the

same. The amplitude can be controlled by pulling out the string-

mass combo to a certain angle as measured by the protractor.

- Mass versus period. Take a period measurement for 5 to 10

different masses while keeping the amplitude and length

constant.

- Length versus period. Take a period measurement for 5 to 10

different lengths while keeping the mass and amplitude constant.

Length is measured from the pivot point to the center of mass.

(b) 1 point for stating that no correlation is expected for amplitude and

mass variations.

1 point for a nonlinear relationship is expected for length and

period:

(c) 1 point. Beyond the usual random errors of measurement (especially

when using a timer but minimized by taking the median value of a

few trials each time), one predictable deviation is in the amplitude

investigation. Pendulums actually behave as simple harmonic

oscillators only under the conditions of small angles (small enough

that $\sin \theta$ is approximately $\theta)$. A large enough amplitude will

require the pendulum to oscillate at larger angles. This means that at

large amplitudes, one can predict the results to deviate from the

expected as the gravitational force no longer acts as a simple

restorative force.

One other possible source of systematic error would be in the mass

investigation. As various masses are swapped out on a fixed length

of string, the students may inadvertently be changing the length of

the string when adding different-sized masses. The length of the

pendulum is from the pivot point to the center of mass. If the

students do not compensate for this by shortening the string when

adding larger masses, they may see an artificial relationship at

higher masses in their graph of mass versus period.

(d) 2 points. To obtain a linear graph, you must plot $T^{2}$ versus L :

The slope of this plot would then be equal to $4 \pi^{2} / g$. Alternatively,

you could plot T versus $\sqrt{L}$.

(e) 1 point for graph, including labels and units.

1 point for the line of best fit.

(f) 1 point. The measured slope is $4 \mathrm{~s}^{2} / \mathrm{m}$. Setting this equal to $4 \pi^{2} / g$,

from part (d), and solving for g gives:

For measuring the mass of at least one cylinder with the digital scale

1 point

For measuring the period of oscillation of the cylinder-spring system with the stopwatch

1 point

For a procedure that indicates that the cylinder hung on the spring should be set into oscillatory motion

1 point

For a procedure that indicates a method to reduce experimental uncertainty

1 point

Accept one of the following:

- For using multiple masses

- For doing multiple trials with a single mass

- For measuring multiple oscillations and dividing by the number of oscillations

Example Response
Place a cylinder on the digital scale and record the mass. Hang the cylinder from the spring and pull the cylinder down a small distance so that the spring is stretched. Release the cylinder. Use the stopwatch to measure the amount of time necessary for the cylinder to complete ten full cycles (from maximum stretch length back to maximum stretch length). Repeat the procedure for cylinders of different masses.

Total for part (a) 4 points

(i) For listing quantities that can be measured with a stopwatch and a digital scale and could be plotted to produce a linear graph whose slope can be used to determine k
Accept one of the following:

- m vs. $T^{2}$

- $T^{2}$ vs. m

- $4 \pi^{2} m$ vs. $T^{2}$

- $T^{2}$ vs. $4 \pi^{2} m$

- $\frac{T^{2}}{4 \pi^{2}}$ vs. m

- m vs. $\frac{T^{2}}{4 \pi^{2}}$

- T vs. $\sqrt{m}$

- $\sqrt{m}$ vs. T

- $\frac{T}{2 \pi}$ vs. $\sqrt{m}$

- $\sqrt{m}$ vs. $\frac{T}{2 \pi}$

- T vs. $2 \pi \sqrt{m}$

- $2 \pi \sqrt{m}$ vs. T

Scoring Note: This point may be earned for any of the bullets above substituting $\frac{1}{f}$ for T.
Example Response
Vertical axis: $m \quad$ Horizontal axis: $T^{2}$

Example Response
Plotting the mass as a function of the period-squared would result in a graph whose slope could be used to find k by using the equation for the period of an oscillating cylinder-spring system.

Graph

Slope

k

m vs. $T^{2}$

slope $=\frac{k}{4 \pi^{2}}$

k=( slope $) 4 \pi^{2}$$T^{2}$ vs. m

slope $=\frac{4 \pi^{2}}{k}$$k=\frac{4 \pi^{2}}{\text { slope }}$$4 \pi^{2} m$ vs. $T^{2}$

slope =k

k= slope

$T^{2}$ vs. $4 \pi^{2} m$

slope $=\frac{1}{k}$$k=\frac{1}{\text { slope }}$$\frac{T^{2}}{4 \pi^{2}}$ vs. m

slope $=\frac{1}{k}$$k=\frac{1}{\text { slope }}$

m vs. $\frac{T^{2}}{4 \pi^{2}}$

slope =k

k= slope

T vs. $\sqrt{m}$

slope $=\sqrt{\frac{4 \pi^{2}}{k}}$$k=\frac{4 \pi^{2}}{\text { slope }^{2}}$$\sqrt{m}$ vs. T

slope $=\sqrt{\frac{k}{4 \pi^{2}}}$

k= slope $^{2} \times 4 \pi^{2}$$\frac{T}{2 \pi}$ vs. $\sqrt{m}$

slope $=\sqrt{\frac{1}{k}}$$k=\frac{1}{\text { slope }^{2}}$$\sqrt{m}$ vs. $\frac{T}{2 \pi}$

slope $=\sqrt{k}$

k= slope $^{2}$

T vs. $2 \pi \sqrt{m}$

slope $=\sqrt{\frac{1}{k}}$$k=\frac{1}{\text { slope }^{2}}$$2 \pi \sqrt{m}$ vs. T

slope $=\sqrt{k}$

k= slope $^{2}$

Total for part (b) 2 points

LO 6.D.3.4, SP 1.2; LO 6.D.4.2, SP 2.2 2 points

Calculate the length L of the tube.

For using $\lambda=v / f$

1 point

$\lambda=(340 \mathrm{~m} / \mathrm{s}) /(512 \mathrm{~Hz})=0.66 \mathrm{~m}$

For a length that is half of the calculated wavelength, with units

1 point

$L=\lambda / 2=0.33 \mathrm{~m}$

LO 6.A.1.2, SP 1.2; LO 6.D.3.2, SP 6.4; LO 6.D.3.4, SP 1.2; LO 6.D.4.2, SP 2.2 3 points

The column of air in the tube is still resonating at its fundamental frequency. On the axes below, sketch a graph of the maximum speed of air molecules as they oscillate in the tube, as a function of position x, from x=0 (left end of tube) to x=L (right end of tube). (Ignore random thermal motion of the air molecules.)

For a curve with a node (zero) at L / 2

1 point

For a curve with maxima at 0, L, and no other points

1 point

For a nonhorizontal curve that is symmetric around L / 2 and nonnegative everywhere

1 point

LO 6.D.3.4, SP 1.2; LO 6.D.4.2, SP 2.2) 2 points

The right end of the tube is now capped shut, and the tube is placed in a chamber that is filled with another gas in which the speed of sound is 1005 m/s. Calculate the new fundamental frequency of the tube.

Correct answer: 757 Hz

For an indication that the fundamental wavelength is 4 L

1 point

For substituting the new sound speed in $v=\lambda f$

1 point

For an explanation that indicates that the maximum kinetic energy and maximum potential 1 point energy are the same due to energy conservation
Scoring Note: This point may be earned for only stating "conservation of energy."
Example Response
The maximum kinetic energy and maximum potential energy of the car-spring system are both 4 J, because energy is conserved in this system.

Total for part (a) 1 point

For using the equation for frequency or period in a ratio 1 point
Example Responses

Scoring Note: Simplified versions of the above ratios also earn this point.

For substituting the total mass $4 m_{0}$ into the correct ratio: $\frac{f_{2}}{f_{1}}$ or $\frac{T_{1}}{T_{2}}$ 1 point

Example Response

Total for part (b) 2 points