AP Physics 1- 7.3 Representing and Analyzing SHM - Exam Style questions - FRQs- New Syllabus
Representing and Analyzing SHM AP Physics 1 FRQ
Unit 7: Oscillations
Weightage : 10-15%
Question
A student hangs a spring of unknown spring constant \( k \) vertically by attaching one end to a stand, as shown in Figure \( 1 \). The other end of the spring has a small loop from which small cylinders can be hung. In addition to the spring, the student has access only to a variety of cylinders of unknown masses, a stopwatch, and a digital scale.
In a different experiment, the student attaches one end of a spring to a force sensor that is attached to a wall. The other end of the spring is attached to a cart with mass \( m = 0.25\ \mathrm{kg} \). The student places a motion detector to the right of the cart, as shown in Figure \( 2 \), and pulls the cart to the right a small distance so that the spring is stretched. The student releases the cart from rest, and the cart-spring system oscillates.
Most-appropriate topic codes (AP Physics 1):
• Topic \( 7.3 \) — Representing and Analyzing SHM (Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \))
• Topic \( 7.4 \) — Energy of Simple Harmonic Oscillators (Part \( \mathrm{(c)(i)} \))
• Topic \( 4.2 \) — Change in Momentum and Impulse (Part \( \mathrm{(c)(ii)} \), Part \( \mathrm{(c)(iii)} \))
• Topic \( 3.1 \) — Translational Kinetic Energy (Part \( \mathrm{(c)(i)} \))
▶️ Answer/Explanation
(a)
One valid method is to use the period of oscillation of the spring-cylinder system for several different hanging masses.
| Quantity to Be Measured | Symbol for Quantity | Equipment for Measurement |
|---|---|---|
| Mass of a cylinder | \( m \) | Digital scale |
| Time for \( N \) oscillations | \( t_N \) | Stopwatch |
| Number of oscillations counted | \( N \) | Direct count |
Procedure:
First, choose one cylinder and use the digital scale to measure its mass \( m \). Hang that cylinder from the spring. Pull it down a small distance and release it so that it oscillates vertically.
Use the stopwatch to measure the total time \( t_N \) for \( N \) complete oscillations. Then calculate the period using
\( T = \dfrac{t_N}{N} \)
Repeat this process for several cylinders of different masses, recording a set of \( (m,T) \) values.
To reduce uncertainty, time many oscillations instead of just one, repeat the timing trials for each mass, and average the measured values of \( T \). Also keep the oscillations small so the motion stays close to simple harmonic motion.
(b)(i)
One valid choice is:
Vertical axis: \( m \)
Horizontal axis: \( T^2 \)
Since
\( T = 2\pi \sqrt{\dfrac{m}{k}} \)
squaring gives
\( T^2 = \dfrac{4\pi^2 m}{k} \)
Rearranging,
\( m = \left(\dfrac{k}{4\pi^2}\right) T^2 \)
so this graph is linear.
(b)(ii)
For a graph of \( m \) versus \( T^2 \), the slope is
\( \text{slope} = \dfrac{k}{4\pi^2} \)
Therefore,
\( k = (\text{slope})\,4\pi^2 \)
So the spring constant is found by multiplying the slope of the best-fit line by \( 4\pi^2 \).
(c)(i)
Use
\( K = \dfrac{1}{2}mv^2 \)
From the velocity-time graph:
\( v_i \approx 0.30\ \mathrm{m/s} \) at \( t = 0.5\ \mathrm{s} \)
\( v_f \approx 0 \) at \( t = 2.0\ \mathrm{s} \)
\( \Delta K = K_f – K_i \)
\( \Delta K = \dfrac{1}{2}(0.25)(0)^2 – \dfrac{1}{2}(0.25)(0.30)^2 \)
\( \Delta K = 0 – 0.01125\ \mathrm{J} \)
\( \boxed{\Delta K \approx -0.0113\ \mathrm{J}} \)
(c)(ii)
The change in momentum is approximately
\( \boxed{0\ \mathrm{kg \cdot m/s}} \)
The change in momentum equals the impulse, which is the area under the force-time graph:
\( \Delta p = \int F\,dt \)
From \( t = 0.5\ \mathrm{s} \) to \( t = 2.5\ \mathrm{s} \), the negative area and positive area under the curve are approximately equal in magnitude, so they cancel. Therefore the net impulse, and thus the change in momentum, is about zero.
(c)(iii)
Yes, the velocity-time graph confirms this estimate.
At \( t = 0.5\ \mathrm{s} \), the velocity is about \( 0.30\ \mathrm{m/s} \), and at \( t = 2.5\ \mathrm{s} \), the velocity is also about \( 0.30\ \mathrm{m/s} \).
Since momentum is
\( p = mv \)
and the mass of the cart is the same at both times, the momentum is the same at both times. Thus,
\( \Delta p = 0 \)
which agrees with the estimate from the force-time graph.