AP Physics 1 7.2: Frequency and Period of SHM

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