AP Physics C Mechanics Unit 7 Oscillations

[Maximum number: 2]

Mech.2.
A block of mass 2 M rests on a horizontal, frictionless table and is attached to a relaxed spring, as shown in the figure above. The spring is nonlinear and exerts a force $F(x)=-B x^{3}$ , where B is a positive constant and x is the displacement from equilibrium for the spring. A block of mass 3 M and initial speed $v_{0}$ is moving to the left as shown.

Do the two blocks, which remain stuck together and attached to the spring, exhibit simple harmonic motion after the collision?

Yes No
Justify your answer.

Figure 1

Figure 2

2 points
For selecting the correct answer "No," with a reasonable attempt at a justification If the incorrect selection is made, no points are earned for the justification. If the correct answer is selected without any justification, the point is not earned for the selection.
For an indication that the spring does not apply a linear force and that simple 1 point harmonic motion is the resulting motion of a linear restoring force $\left(a=-\frac{k}{m} \Delta x\right)$
Example:
Because the blocks are sticking together and are attached to the spring, the spring will apply a restoring force to the blocks. However, because the restoring force exerted by a nonlinear spring is not proportional to the blocks' displacement from equilibrium, the blocks do not exhibit simple harmonic motion.

[Maximum number: 2]

Block A and Block B of masses m and 3 m, respectively, are arranged in a setup consisting of an ideal spring with spring constant k and a horizontal surface. Friction between the surface and the blocks is negligible except in a region of length D, where the coefficient of kinetic friction between Block A and the surface is $\mu$ . Block B is attached to a string of length $\ell$ and negligible mass, as shown in Figure 1. Block A is held against the spring, compressing the spring a distance $x_{\mathrm{c}}$ .

At time t=0, Block A is located at position $x=x_{0}$ and is released from rest. After the block is released, the following occurs.

- At time $t=t_{1}$ , Block A is at $x=x_{1}$ after traveling a distance $x_{\mathrm{c}}$ . Block A moves with speed v, and the spring is at its equilibrium position.

- At time $t=t_{2}$ , the left side of Block A is at $x=x_{2}$ after passing through a distance D across the region with nonnegligible friction.

- At time $t=t_{3}$ , Block A is at $x=x_{3}$ and Block A collides with and sticks to Block B.

Indicate how the new frequency of oscillation $f_{2 \ell}$ of the system on the new string of length $2 \ell$ will compare to the frequency of oscillation $f_{\ell}$ from the original procedure. $f_{2 \ell}>f_{\ell}$ $f_{2 \ell}<f_{\ell}$ $f_{2 \ell}=f_{\ell}$

Briefly justify your answer.

Figure 1

For selecting $f_{2 \ell}<f_{\ell}$ with an attempt at a relevant justification 1 point
For correctly applying an equation that relates the length of a pendulum to the period or 1 point frequency of the pendulum
Example Response
The period of a pendulum is calculated by using $T=2 \pi \sqrt{\frac{l}{g}}$ . Therefore, as the length is increased, the period will also increase. Because frequency and period are inversely related, an increase in period will result in a decrease in frequency.

Total for part (c) for question 115 points

[Maximum number: 11]

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 7 Oscillations question 3 — AP Physics C: Mechanics

(a)

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

[ 4 ]

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.

[ 2 ]

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)

Calculate the value of k.

Experiment 2. The block and spring arrangement is now placed on a rough surface, as shown below. The block is displaced so that the spring is compressed a distance d and released from rest.

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

[ 2 ]

2 points
For a correct relationship between the period and the spring constant 1 point

T=2 \pi \sqrt{\frac{m}{k}}

For substituting correct values from previous parts into a correct expression 1 point

\begin{aligned} & k=\frac{4 \pi^{2} m}{T^{2}}=\frac{\left(4 \pi^{2}\right)(0.30 \mathrm{~kg})}{(0.70 \mathrm{~s})^{2}} \\ & k=24 \mathrm{~N} / \mathrm{m} \end{aligned}

Alternate solution \#1
For a correct expression relating angular frequency and the spring constant 1 point
$\omega=\sqrt{\frac{k}{m}}$
For substituting correct values from previous parts into a correct expression 1 point

Alternate solution \#2
For a correct statement of the conservation of energy, applied to the position of maximum displacement and the equilibrium position $\frac{1}{2} k x_{\text {max }}^{2}=\frac{1}{2} m v_{\text {max }}^{2}$
For substituting correct values from previous parts into a correct expression 1 point

(d)

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 (d) — AP Physics C: Mechanics

Mech. 2.

[ 3 ]

[Maximum number: 13]

A student makes a torsional pendulum by suspending a uniform disk of mass M and radius R from a light wire with torsion constant $\kappa$ that is attached to the center of the disk as shown in Figure 1. The rotational inertia of the disk is given by $I=\frac{1}{2} M R^{2}$ . The student conducts an investigation to determine the relationship between the period of oscillation T of the torsional pendulum and the number N of identical disks that are suspended from the wire.
The student starts with a single disk. Holding the disk at a small initial angular displacement $\theta_{0}$ from the untwisted position, the student releases the disk from rest and the pendulum oscillates. The student records the period of oscillation for a single disk. An additional identical disk is attached, as shown in Figure 2, and the procedure is repeated for N=2 disks. This procedure is repeated through N=10 identical disks. Assume the disks move together as one system.

(a)

Using $T=2 \pi \sqrt{\frac{I}{\kappa}}$ , derive an expression for T as a function of N. Express your answer in terms of $M, R, \kappa, N$ , and physical constants, as appropriate.

[ 2 ]

For indicating the rotational inertia is the sum of the rotational inertia for all the stacked disks 1 point
Example Response

\begin{aligned} I_{\mathrm{eq}} & =\sum_{i=1}^{N} I_{i} \\ I_{\mathrm{eq}} & =N\left(\frac{1}{2} M R^{2}\right) \end{aligned}

For an expression for the period consistent with the previous rotational inertia expression 1 point
Example Response

T=2 \pi \sqrt{\frac{N\left(\frac{1}{2} M R^{2}\right)}{\kappa}}

Example Solution

\begin{aligned} & I_{\mathrm{eq}}=\sum_{i=1}^{N} I_{i} \\ & I_{\mathrm{eq}}=N\left(\frac{1}{2} M R^{2}\right) \\ & T=2 \pi \sqrt{\frac{I}{\kappa}} \\ & T=2 \pi \sqrt{\frac{N\left(M R^{2}\right)}{2 \kappa}} \end{aligned}

(b)

The potential energy stored in the torsional pendulum when the disks are displaced is $U=\frac{1}{2} \kappa(\Delta \theta)^{2}$ . On the following axes, sketch a graph of the maximum kinetic energy $K_{\text {max }}$ of the torsional pendulum as a function of N for $N \geq 1$ .

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

[ 2 ]

(c)

The student plots the data for T as a function of $\sqrt{N}$ , as shown.

[ 6 ]

(i)

Draw the best-fit line for the data.

[ 1 ]

(ii)

The student previously determined that the radius of a disk is $R=0.2 \mathrm{~m}$ and found that $\kappa=1.6 \mathrm{~N} \cdot \mathrm{~m}$ . Using the graph, calculate the mass M of a single disk.

[ 3 ]

(ii) For using two points on the line to calculate the slope 1 point
Example Response
slope $=\frac{T_{2}-T_{1}}{\sqrt{N_{2}}-\sqrt{N_{1}}}$
slope $=\frac{0.4 \mathrm{~s}-0.2 \mathrm{~s}}{1.5-0.8}=0.29 \mathrm{~s}$

Scoring Note: Slope values may range from. 22 s to. 33 s.
For correctly relating the slope to the period of the torsional pendulum consistent with 1 point part (a)
Example Response
$T=2 \pi \sqrt{\frac{N I}{\kappa}}$
→ slope $=2 \pi \sqrt{\frac{I}{\kappa}}$
For substituting the slope into the equation to determine the mass of the disk 1 point
Example Response
$\rightarrow M=\frac{\kappa(\text { slope })^{2}}{2 \pi^{2} R^{2}}$
Example Solution

\begin{aligned} & T=2 \pi \sqrt{\frac{N I}{\kappa}} \\ & \rightarrow \text { slope }=2 \pi \sqrt{\frac{I}{\kappa}} \\ & (\text { slope })^{2}=4 \pi^{2} \frac{I}{\kappa} \\ & (\text { slope })^{2}=4 \pi^{2} \frac{\left(\frac{1}{2} M R^{2}\right)}{\kappa} \\ & \rightarrow M=\frac{\kappa(\text { slope })^{2}}{2 \pi^{2} R^{2}} \\ & M=\frac{(1.6 \mathrm{~N} \cdot \mathrm{~m})\left(\frac{0.4 \mathrm{~s}-0.2 \mathrm{~s}}{1.5-0.8}\right)^{2}}{2 \pi^{2}(0.2 \mathrm{~m})^{2}} \\ & \therefore M=0.17 \mathrm{~kg} \end{aligned}

(iii)

The student finds that the value given by the manufacturer for the mass of the disk is less than the value determined experimentally in part (c)(ii). Determine a single source of experimental error that could result in the observed difference in the value of M. Justify your answer.

[ 2 ]

For indicating a source of error that could correctly explain the observed difference with an attempt at a relevant justification
For a correct justification that links the source of experimental error to the larger 1 point experimental value for M. Accept one of the following:

- Experimental uncertainties for the period, for example the period is measured to be larger

- The mass is more concentrated to the edge, or some other distribution that results in a larger rotational inertia

Scoring Note: Responses that indicate the given $\kappa$ is too large or the given radius is too small may earn the second point.
Example Solution

The experimental value of the mass could be too large because the period measured by the student is too large.

OR

The experimental value of the mass could be too large because the mass of the disk is concentrated at the edge of the disk, causing the rotational inertia of the disk to be larger than that used to determine the experimental value of mass.

Total for part (c)

6 points

(d)

The student repeats the experiment, but now the disks have a density that varies as a function of the radius of the disk according to $\rho=0.3 r$ .

[ 3 ]

(i)

For indicating that the slope would be greater with an attempt at a relevant justification

1 point

For indicating that using disks with densities that increase with r will increase the rotational
inertia

1 point

For indicating the functional relationship between slope and rotational inertia: $\sqrt{I} \propto$ slope

1 point

Example Solution

The disks with a density that increases towards the edge of the disk will have a greater proportion of their mass farther from the axis of rotation, so their rotational inertia will be
larger than that of a uniform disk. Therefore, the slope of the line will be greater because
the slope is proportional to $\sqrt{I}$ .

(i)

Would the slope of the best-fit line for this new data be greater than, less than, or the same as the slope of the best-fit line in part (c)(i) ?

greater than less than the same as

Justify your answer.

[ 3 ]

For indicating that the slope would be greater with an attempt at a relevant justification

1 point

For indicating that using disks with densities that increase with r will increase the rotational
inertia

1 point

For indicating the functional relationship between slope and rotational inertia: $\sqrt{I} \propto$ slope

1 point

Example Solution

The disks with a density that increases towards the edge of the disk will have a greater proportion of their mass farther from the axis of rotation, so their rotational inertia will be
larger than that of a uniform disk. Therefore, the slope of the line will be greater because
the slope is proportional to $\sqrt{I}$ .

AP Physics C Mechanics Unit 7: Oscillations

Covered syllabus points

Covered syllabus points

7 Oscillations question 1

7 Oscillations question 2

7 Oscillations question 3

Question (a)

Question (b)

Question (c)

Question (d)

7 Oscillations question 4

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (iii)

Question (d)

Question (i)