Ten sense-making tasks on simple harmonic motion: springs, pendulums, graphs, and energy. Explain your reasoning for every answer. The explanation counts as much as the answer.
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Each block is attached to a spring on a frictionless horizontal surface. Each block is pulled out to the amplitude $A$ shown and released from rest. The mass of each block and the spring constant of each spring are given.
Rank the oscillators by their period, from longest to shortest. If any are equal, show the tie. If the ranking cannot be determined, say so.
Longest 1 ____ 2 ____ 3 ____ 4 ____ 5 ____ 6 ____ Shortest
Explain your reasoning.
Back to topA simple pendulum of length $L$ with a bob of mass $m$ swings back and forth with a small angular amplitude $\theta$. Three students are trying to figure out how to make the pendulum take longer to complete one swing.
Ava: "Use a heavier bob. Gravity pulls harder on a heavier bob, so it's harder to get it moving and it will swing more slowly."
Ben: "Pull it back farther before releasing it. With a bigger amplitude the bob has a longer path to travel, so each swing takes more time."
Cam: "Only the length matters here. Make the string longer. If you make it four times longer, the period doubles."
Which student(s), if any, do you agree with? Explain what is right or wrong about each statement. Is there anything else (besides what the students mentioned) that would change the period?
Back to topA block on a horizontal spring oscillates on a frictionless surface between $x = -A$ and $x = +A$. A student draws the net force (red) and velocity (green) on the block at three positions, while the block is moving to the right through $x = 0$:
Student's explanation: "At $x = 0$ the block is moving fastest, so that's where the net force on it is biggest. At the ends the block stops, so the net force there is zero."
What, if anything, is wrong with the student's diagram and explanation? If something is wrong, explain the error and describe the correct diagram. If nothing is wrong, explain why it is correct.
Back to topA block on a spring and a simple pendulum are both oscillating with small amplitudes on Earth. For each change, choose what happens to the period of that oscillator:
A. increases B. decreases C. stays the same D. cannot be determined
| Change | Choice |
|---|---|
| (a) Spring: the mass of the block is quadrupled | |
| (b) Spring: the amplitude is doubled | |
| (c) Spring: the spring is replaced with a stiffer spring | |
| (d) Spring: the whole setup is taken to the Moon | |
| (e) Pendulum: the mass of the bob is doubled | |
| (f) Pendulum: the string is made longer | |
| (g) Pendulum: the whole setup is taken to the Moon | |
| (h) Pendulum: the bob mass is doubled and the string is made shorter |
Explain the rule you used. Why do (d) and (g) have different answers?
Back to topTwo identical 2.0 kg blocks are attached to identical springs ($k = 200\ \text{N/m}$) on a frictionless surface. Block 1 is pulled 0.10 m from equilibrium and released from rest. Block 2 is pulled 0.20 m from equilibrium and released from rest.
For each quantity, state whether it is greater for Block 1, greater for Block 2, or the same for both. If it's different, give the ratio.
| Quantity | Answer |
|---|---|
| (a) Period of oscillation | |
| (b) Maximum force exerted by the spring | |
| (c) Maximum acceleration | |
| (d) Total mechanical energy of the block-spring system | |
| (e) Maximum speed |
Explain each answer. Then check (e) by finding the actual maximum speed of each block.
Back to topIn Setup 1, a 0.50 kg block is attached to a spring ($k = 50\ \text{N/m}$) on a frictionless horizontal surface and set oscillating. In Setup 2, the same spring and block hang vertically from the ceiling, and the block is pulled down and released so it bounces up and down.
The equation below describes the energy of an oscillating system at one instant:
$$\tfrac{1}{2}(100\ \text{N/m})(0.20\ \text{m})^2 = \tfrac{1}{2}(3.0\ \text{kg})\,v^2 + \tfrac{1}{2}(100\ \text{N/m})(0.10\ \text{m})^2$$
The graph shows the position of a block oscillating on a spring.
A 0.20 kg block on a spring ($k = 20\ \text{N/m}$) slides on a frictionless surface. It is pulled to $x = +0.10\ \text{m}$ and released from rest. How long does it take the block to reach $x = -0.10\ \text{m}$ for the first time?
Student solution:
$T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{0.20\ \text{kg}}{20\ \text{N/m}}} = 2\pi(0.10\ \text{s}) = 0.63\ \text{s}$
"Going from $+0.10\ \text{m}$ to $-0.10\ \text{m}$ is one swing in one direction, which is a quarter of a cycle."
$t = \tfrac{1}{4}T = \tfrac{1}{4}(0.63\ \text{s}) = 0.16\ \text{s}$
Identify the error in the solution. Explain the physics of why it's an error, then find the correct time. How long would it take to go from $x = +0.10\ \text{m}$ to $x = 0\ \text{m}$?
Back to topA 2.0 kg block on a frictionless surface is attached to a spring ($k = 200\ \text{N/m}$). It is pulled 0.10 m from equilibrium and released from rest.
Riley writes: $a = \dfrac{kx}{m} = \dfrac{(200\ \text{N/m})(0.10\ \text{m})}{2.0\ \text{kg}} = 10\ \text{m/s}^2$. Then $v^2 = v_0^2 + 2a\Delta x = 0 + 2(10\ \text{m/s}^2)(0.10\ \text{m})$, so the block reaches equilibrium at $v = 1.4\ \text{m/s}$.