AP Physics C Mechanics 7.3 Representing SHM

[Maximum number: 9]

Experiment 1. A block of mass 0.30 kg is placed on a frictionless table and is attached to one end of a horizontal spring of spring constant k, as shown above. The other end of the spring is attached to a fixed wall. The block is set into oscillatory motion by stretching the spring and releasing the block from rest at time t=0. A motion detector is used to record the position of the block as it oscillates. The resulting graph of velocity v versus time t is shown below. The positive direction for all quantities is to the right.

Figure for Question 1 — AP Physics C: Mechanics

(a)

Determine the equation for v(t), including numerical values for all constants.

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4 points
For writing a correct trigonometric equation for velocity as a function of time, including 1 point the negative sign

v(t)=-v_{\max } \sin (\omega t)=-v_{\max } \sin (2 \pi t / T)

For using $\omega=2 \pi f$ or $\omega=\frac{2 \pi}{T}$ to solve for $\omega$ 1 point
For using the correct period of 0.70 s from the graph 1 point

\omega=\frac{2 \pi}{0.70 \mathrm{~s}}=9.0 \mathrm{rad} / \mathrm{s}

For using the correct value of the maximum speed from the graph (acceptable range of 1 point values for $v_{\text {max }}: 0.15 \mathrm{~m} / \mathrm{s}$ to $0.17 \mathrm{~m} / \mathrm{s}$ ) $v(t)=(-0.16) \sin (9.0 t)$

Note: One point is deducted if incorrect phase shift $\phi$ is used. Full credit is awarded for a correct answer with no work shown. Students are also given credit if the value of k from part (c) is used to calculate $\omega$ using $\omega=\sqrt{k / m}$ .

(b)

Given that the equilibrium position is at x=0, determine the equation for x(t), including numerical values for all constants.

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2 points
Take the integral of the velocity determined in part (a)

x(t)=\int v(t) d t=\int(-0.16 \mathrm{~m} / \mathrm{s}) \sin ((9.0 \mathrm{rad} / \mathrm{s}) t) d t

For a correct trigonometric expression consistent with integrating the answer from part (a)
For a correct $x_{\text {max }}$ consistent with the integrating the answer from part (a) 1 point

\begin{aligned} & x_{\max }=(0.16 \mathrm{~m} / \mathrm{s}) /(9.0 \mathrm{rad} / \mathrm{s})=0.018 \mathrm{~m} \\ & x(t)=(0.018) \cos (9.0 t) \end{aligned}

Alternate solution
For solving for a maximum displacement consistent with the answer from part (a) $v_{\text {max }}=x_{\text {max }} \omega$

\begin{aligned} & x_{\max }=\frac{v_{\max }}{\omega}=\frac{(0.16 \mathrm{~m} / \mathrm{s})}{(9.0 \mathrm{rad} / \mathrm{s})} \\ & x_{\max }=0.018 \mathrm{~m} \end{aligned}

For a correct trigonometric expression consistent with the answer from part (a) $x(t)=(0.018) \cos (9.0 t)$
Note: Full credit is awarded for a correct answer with no work shown. One earned point is deducted for incorrect initial conditions (e.g., subtracting a constant from the cosine function).

(c)

Draw a sketch of v versus t in this case. Assume that there is a negligible change in the period and that the positive direction is still to the right.

Figure for Question (c) — AP Physics C: Mechanics

Mech. 2.

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AP Physics C: Mechanics 7.3 Representing and Analyzing Shm Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)