AP Physics 1 Quiz: Frequency And Period Of Shm

What this quiz covers

This quiz focuses on Frequency And Period Of Shm, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. It increases because the weight is smaller.
2. It decreases because the mass accelerates less.
3. It is unchanged because $T=2\pi\sqrt{m/k}$ does not involve $g$. (correct answer)
4. It changes only if the amplitude changes.

Explanation: This question tests understanding of the period in simple harmonic motion for a horizontal mass-spring system under varying gravity. The period T = 2π√(m/k) relies solely on mass and spring constant, independent of gravity since the surface is horizontal. Qualitatively, gravity affects vertical equilibrium but not the horizontal oscillation dynamics. Therefore, reducing gravity leaves the period unchanged. Choice A is a distractor that assumes the period increases due to smaller weight, mistakenly applying vertical oscillator logic to a horizontal setup. Differentiate between horizontal and vertical oscillators, noting gravity's role only in the latter for equilibrium position.

Question 2

1. It doubles because the mass doubled.
2. It halves because the spring is stiffer.
3. It is unchanged because $T=2\pi\sqrt{m/k}$ and the ratio $m/k$ is the same. (correct answer)
4. It increases because the amplitude will increase when $k$ increases.

Explanation: This question examines what happens to period when both mass and spring constant change proportionally. The period of a mass-spring oscillator is T = 2π√(m/k), which depends on the ratio m/k. When both m and k are doubled, the ratio m/k remains unchanged: (2m)/(2k) = m/k. Therefore, the period T' = 2π√(2m/2k) = 2π√(m/k) = T remains the same. Choice A incorrectly focuses only on the mass doubling without considering the spring constant change. The strategy is to look at how parameters appear in the period formula: when they appear as a ratio, proportional changes cancel out.

Question 3

1. $9T$ because a stiffer spring makes the force larger.
2. $3T$ because the block moves three times faster.
3. $\tfrac{T}{3}$ because $T\propto 1/\sqrt{k}$. (correct answer)
4. $\tfrac{T}{9}$ because $T\propto 1/k$.

Explanation: This question tests how period depends on spring constant in simple harmonic motion. The period of a mass-spring oscillator is T = 2π√(m/k), so period is inversely proportional to the square root of spring constant. When k increases to 9k, the new period becomes T' = 2π√(m/9k) = (1/3)·2π√(m/k) = T/3. Choice D incorrectly assumes inverse proportionality (T ∝ 1/k) rather than inverse square root, which would give T/9. The key is remembering that stiffer springs produce larger restoring forces and thus faster oscillations, with period decreasing as 1/√k.

Question 4

1. It increases if $g$ increases because the weight increases.
2. It decreases if $g$ decreases because the amplitude becomes larger.
3. It is unchanged because horizontal mass–spring frequency depends on $m$ and $k$, not $g$. (correct answer)
4. It changes because frequency equals maximum speed divided by amplitude.

Explanation: This question tests understanding of what affects frequency in horizontal mass-spring motion. For a mass-spring system oscillating horizontally, the angular frequency is ω = √(k/m), which depends only on spring constant k and mass m, not on gravitational acceleration g. Moving to a planet with different g doesn't affect horizontal oscillations because gravity acts perpendicular to the motion and doesn't contribute to the restoring force. Choice A incorrectly thinks increased weight affects horizontal motion, confusing this with vertical oscillations where gravity shifts the equilibrium position. The key insight is that horizontal SHM depends only on the spring's restoring force, while vertical SHM includes gravity in determining equilibrium but not frequency.

Question 5

1. It decreases because weaker gravity means less time to fall.
2. It increases because $T\propto 1/\sqrt{g}$. (correct answer)
3. It is unchanged because the length is the same.
4. It depends on the bob’s mass, which is unchanged.

Explanation: This question tests how gravitational acceleration affects pendulum period. For a simple pendulum, the period is T = 2π√(L/g), showing that period is inversely proportional to the square root of g. When g decreases (weaker gravity), the denominator becomes smaller, making the overall expression larger, so period increases. Choice A incorrectly reasons that weaker gravity means less time to fall, not recognizing that weaker gravity actually means smaller restoring force and thus slower oscillation. The key insight is that on planets with weaker gravity, pendulums swing more slowly because the component of weight providing the restoring force is reduced.

Question 6

1. It increases because the bob travels farther each cycle.
2. It decreases because the maximum speed increases.
3. It changes because the restoring force depends on amplitude.
4. It remains approximately the same because $T$ depends mainly on $L$ and $g$. (correct answer)

Explanation: This question assesses the period's amplitude independence in simple harmonic motion for a small-angle pendulum. The period T ≈ 2π√(L/g) holds for small amplitudes, depending mainly on length and gravity. Qualitatively, small increases in amplitude slightly alter the motion but keep it approximately harmonic, with negligible period change. Therefore, the period remains approximately the same. Choice C is a distractor that claims it changes due to amplitude-dependent restoring force, which is true only for large angles beyond SHM approximation. For amplitude variations, confirm if conditions remain within SHM limits and apply the approximation accordingly.

Question 7

1. It doubles because the inertia is larger.
2. It is unchanged because frequency depends on amplitude, not mass.
3. It increases by a factor of $2$ because the maximum speed decreases.
4. It decreases by a factor of $2$ because $f\propto 1/\sqrt{m}$. (correct answer)

Explanation: This question examines the frequency dependence in simple harmonic motion for a mass-spring oscillator. The frequency f = (1/2π)√(k/m) decreases with increasing mass, as greater inertia slows the response to the restoring force. Qualitatively, replacing m with 4m quadruples the inertia, reducing frequency by √(1/4) = 1/2. Thus, the new frequency is half the original. Choice A is a distractor that incorrectly states the period doubles due to inertia, but the question asks about frequency, not period. For mass changes, remember frequency scales inversely with √m and verify the asked quantity—frequency or period.

Question 8

1. It becomes $4$ times larger because the bob moves through a longer arc.
2. It doubles because $T\propto \sqrt{L}$. (correct answer)
3. It halves because the restoring force increases with length.
4. It stays the same because period depends only on mass.

Explanation: This question evaluates knowledge of the period in simple harmonic motion for a simple pendulum. The period of a simple pendulum is T = 2π√(L/g), indicating it increases with the square root of length and is independent of mass or amplitude for small angles. Qualitatively, longer length means the bob travels a greater arc but with weaker effective restoring force, leading to a longer period. Therefore, increasing length to 4L multiplies the period by √4 = 2. Choice A is a distractor that wrongly suggests the period quadruples due to the longer arc, confusing distance with the actual dependency on √L. A useful strategy is to memorize the SHM period formulas and identify which variables are truly independent, like mass in pendulums.

Question 9

1. $f_Y = 3f_X$ because the amplitude is three times larger.
2. $f_Y = \tfrac{1}{3}f_X$ because a larger amplitude makes each cycle longer.
3. $f_Y = f_X$ because frequency depends only on $m$ and $k$. (correct answer)
4. $f_Y > f_X$ because the maximum restoring force is larger.

Explanation: This question assesses understanding of the frequency and period in simple harmonic motion for mass-spring systems. The frequency f for a mass-spring system is f = (1/2π) √(k/m), depending on spring constant k and mass m but independent of amplitude. Period T = 1/f is also amplitude-independent, as the oscillation rate is set by the system's stiffness and inertia. Tripling amplitude from A to 3A leaves frequency unchanged, as it does not alter m or k. A common distractor is choice A, which wrongly triples frequency assuming larger amplitude increases speed in a way that shortens cycles, but in SHM, time per cycle remains constant. To solve such problems, recall the standard period formulas for SHM systems and identify which parameters are truly independent of the variable changed.

Question 10

1. The frequency triples. (correct answer)
2. The frequency increases by a factor of 9.
3. The frequency is unchanged because the amplitude is unchanged.
4. The frequency decreases by a factor of 3 because the spring is stiffer.

Explanation: This question assesses understanding of the frequency and period in simple harmonic motion for mass-spring systems. The frequency f of a mass-spring oscillator is f = (1/2π) √(k/m), so it increases with the square root of the spring constant k when mass is constant. Period T = 1/f decreases as k increases, meaning stiffer springs lead to faster oscillations. Replacing k with 9k triples the frequency since √9 = 3, while amplitude remains irrelevant to frequency. A common distractor is choice B, which incorrectly multiplies by 9 linearly instead of taking the square root of the change in k. To solve such problems, recall the standard period formulas for SHM systems and identify which parameters are truly independent of the variable changed.

Question 11

1. It doubles because the block travels twice as far each cycle.
2. It halves because the maximum speed increases.
3. It stays the same because $T$ depends only on $m$ and $k$. (correct answer)
4. It increases because frequency is proportional to amplitude.

Explanation: This question assesses understanding of the period in simple harmonic motion for a mass-spring system. The period of oscillation for a mass-spring system is given by T = 2π√(m/k), showing it depends on the mass and spring constant but not on the amplitude. Qualitatively, increasing the amplitude means the block travels farther, but the restoring force also scales proportionally, keeping the time per cycle constant. Thus, doubling the amplitude does not affect the period, as the motion remains harmonic. A common distractor is choice A, which incorrectly assumes the period doubles because the distance traveled increases, ignoring that speed also increases proportionally. To approach similar problems, always recall that for ideal SHM, period is independent of amplitude and focus on the formula's dependencies.

Question 12

1. $T/2$, because the bob moves faster on a longer arc.
2. $4T$, because the distance traveled each cycle quadruples.
3. $2T$, because $T\propto\sqrt{L}$. (correct answer)
4. $T$, because gravity is unchanged.

Explanation: This question tests understanding of how pendulum length affects period. For small angle oscillations, the period formula T = 2π√(L/g) shows that period is proportional to the square root of length. When length increases from L to 4L, the new period becomes T' = 2π√(4L/g) = 2 × 2π√(L/g) = 2T. While the bob does travel a longer arc with increased length, this alone doesn't determine the period; the key is that the restoring force (component of gravity) becomes weaker relative to the increased arc length. Choice B incorrectly assumes the distance quadruples the period linearly. The strategy is to apply the square root relationship between length and period for all small-angle pendulum problems.

Question 13

1. $f_{16k}=16f_k$ because the restoring force is 16 times larger
2. $f_{16k}=4f_k$ because $f\propto \sqrt{k}$ (correct answer)
3. $f_{16k}=\tfrac{1}{4}f_k$ because the period increases with stiffness
4. $f_{16k}=f_k$ because frequency depends on amplitude, not $k$

Explanation: This question compares frequencies of systems with different spring constants. Since frequency f = (1/2π)√(k/m), when spring constant increases from k to 16k, the frequency becomes f₁₆k = (1/2π)√(16k/m) = 4·(1/2π)√(k/m) = 4fk. The 16-fold increase in spring constant produces only a 4-fold increase in frequency due to the square root relationship. Choice A incorrectly assumes direct proportionality between frequency and spring constant, missing the square root. The strategy for comparing SHM systems is to write the ratio of frequencies using the formula, then simplify: f₂/f₁ = √(k₂/k₁) when masses are equal.

Question 14

1. The larger-amplitude pendulum has a longer period because it travels farther.
2. The larger-amplitude pendulum has a shorter period because its maximum speed is larger.
3. They have the same period because, for small angles, $T$ depends on $L$ and $g$. (correct answer)
4. Their periods differ because the tension force is different at different amplitudes.

Explanation: This question tests whether amplitude affects the period of a simple pendulum. For small angle oscillations, the period T = 2π√(L/g) depends only on length L and gravitational acceleration g, not on amplitude. This is the small angle approximation where sin(θ) ≈ θ, making the restoring force proportional to displacement and thus creating true SHM. While larger amplitude does mean the bob travels farther and reaches higher maximum speed, these effects exactly compensate, leaving period unchanged. Choice A incorrectly focuses only on distance without considering speed variations. The strategy is to recognize that the small angle approximation ensures amplitude independence, which breaks down only for large angles where sin(θ) ≠ θ.

Question 15

1. The period doubles because the block travels twice the distance each cycle
2. The period is unchanged because $T=2\pi\sqrt{m/k}$ (correct answer)
3. The period halves because the maximum speed doubles
4. The period increases because larger amplitude increases the restoring force

Explanation: This question tests understanding of how amplitude affects the period in simple harmonic motion. For a mass-spring system, the period is given by T = 2π√(m/k), which depends only on the mass m and spring constant k, not on the amplitude A. When amplitude doubles, the block does travel twice the distance per cycle, but it also moves proportionally faster at each point, so the time for one complete oscillation remains unchanged. Choice A incorrectly assumes that doubling distance means doubling time, ignoring that speed also changes. The key strategy is to remember that for SHM, period and frequency are independent of amplitude—they depend only on system parameters like mass and spring constant.

Question 16

1. It is four times as large because the length is four times as large.
2. It is half as large because frequency is inversely proportional to $\sqrt{L}$. (correct answer)
3. It is unchanged because frequency depends only on the bob’s mass.
4. It is twice as large because the bob moves faster in a longer arc.

Explanation: This question assesses understanding of the frequency and period in simple harmonic motion for simple pendulums. The period of a simple pendulum for small angles is T = 2π √(L/g), so it depends on length L and gravity g but not on bob mass. Frequency f = 1/T is inversely proportional to √L, meaning longer pendulums oscillate less frequently. Quadrupling L to 4L halves the frequency since √4 = 2 and f ∝ 1/√L. A common distractor is choice A, which incorrectly quadruples the frequency by linearly scaling with length instead of using the square root relationship. To solve such problems, recall the standard period formulas for SHM systems and identify which parameters are truly independent of the variable changed.

Question 17

1. It increases by a factor of $\sqrt{2}$ because heavier bobs swing more slowly.
2. It decreases by a factor of $\sqrt{2}$ because gravity pulls harder.
3. It is unchanged because for small angles $T$ depends on $L$ and $g$ only. (correct answer)
4. It doubles because the mass doubled.

Explanation: This question examines whether pendulum period depends on bob mass. For a simple pendulum with small-angle oscillations, the period is T = 2π√(L/g), which depends only on length L and gravitational acceleration g, not on the bob's mass m. When mass doubles from m to 2m, the gravitational force doubles but so does the inertia, and these effects exactly cancel in the equation of motion. Choice A incorrectly assumes heavier masses swing more slowly, ignoring that the increased weight provides proportionally more restoring force. The strategy is to remember that for pendulums (unlike mass-spring systems), the mass cancels out of the period formula.

Question 18

1. It increases by a factor of $\sqrt{2}$ because there are two springs.
2. It is unchanged because the mass is the same.
3. It decreases by a factor of $\sqrt{2}$ because the effective spring constant is $2k$. (correct answer)
4. It decreases by a factor of 2 because the amplitude is shared by two springs.

Explanation: This question examines how parallel springs affect oscillation period. When two identical springs of constant k support a mass in parallel, they act like a single spring with effective constant keff = 2k because the total restoring force is the sum of forces from both springs. Since period T = 2π√(m/k), with the effective spring constant doubled, the new period becomes T' = 2π√(m/2k) = (1/√2)·2π√(m/k) = T/√2. Choice D incorrectly suggests the amplitude is somehow shared between springs, which misunderstands how parallel springs combine. The strategy is to find the effective spring constant first: springs in parallel add their constants, springs in series add reciprocals.

Question 19

1. It increases because the block’s maximum speed increases.
2. It decreases because the block travels farther each cycle.
3. It is unchanged because frequency is set by $m$ and $k$. (correct answer)
4. It becomes proportional to amplitude because the restoring force is larger.

Explanation: This question examines the relationship between amplitude and frequency in simple harmonic motion. For a mass-spring system, the frequency f = (1/2π)√(k/m) depends only on the spring constant k and mass m, not on the amplitude of oscillation. While increasing amplitude does increase the maximum speed (vmax = Aω where ω is fixed), it also increases the distance traveled, and these effects balance so that the period remains constant. Choice B incorrectly assumes that traveling farther must take more time, not recognizing that speed scales proportionally with distance in SHM. The strategy is to remember that SHM frequency is set by system parameters (m, k for springs; L, g for pendulums), not by how far the oscillator swings.

Question 20

1. It doubles because the block travels twice as far each cycle.
2. It is unchanged because $T$ depends only on $m$ and $k$ for SHM. (correct answer)
3. It halves because the restoring force is larger at larger displacement.
4. It increases because frequency depends on amplitude.

Explanation: This question tests understanding of how period depends on system parameters in simple harmonic motion. For a mass-spring system, the period is given by T = 2π√(m/k), which depends only on the mass m and spring constant k, not on the amplitude A. When amplitude doubles from A to 2A, the block does travel twice as far each cycle, but it also reaches higher maximum speeds proportionally, so the time for one complete oscillation remains unchanged. Choice A incorrectly assumes that traveling farther means taking more time, ignoring that the restoring force (and thus acceleration) scales with displacement in SHM. The key strategy is to remember that for ideal SHM systems (mass-spring and small-angle pendulum), period and frequency are independent of amplitude.