Frequency and Period of SHM - AP Physics 1

A simple pendulum of length $L$ swings with small amplitude. If the bob's mass is increased from $m$ to $2m$, the period is
Answer: unchanged This question examines how frequency and period of a simple pendulum depend on its physical parameters. For a simple pendulum with small amplitude oscillations, the period is T = 2π√(L/g), which remarkably does not depend on the mass of the bob. This occurs because while a heavier bob experiences a larger gravitational force, it also has proportionally more inertia to overcome, and these effects exactly cancel out. The period depends only on the length L and the gravitational field strength g. Choice B incorrectly assumes that doubling the weight doubles the period, failing to recognize that mass cancels out in the pendulum formula. When analyzing pendulum motion, remember that mass independence is a key characteristic that distinguishes pendulums from mass-spring systems.
A simple pendulum of length $L$ swings with small amplitude. The length is reduced to $L/9$ while keeping the same bob. How does the frequency change?
Answer: It increases by a factor of 3 because $f \propto 1/\sqrt{L}$ . This question assesses understanding of the frequency and period in simple harmonic motion for simple pendulums. The frequency f of a simple pendulum is f = (1/2π) √(g/L), so it scales inversely with the square root of length L when g and mass are constant. Period T decreases as L decreases, meaning shorter pendulums oscillate faster. Reducing L to L/9 triples frequency since √(1/9) = 1/3 and f ∝ 1/√L, so f new = 3f old. A common distractor is choice A, which incorrectly decreases frequency by 9 by linearly inverting L 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.
A pendulum of length $L$ oscillates with small amplitude and period $T$ . If the pendulum is moved to a location where $g$ is smaller, how does $T$ change?
Answer: It increases because $T\propto 1/\sqrt{g}$ . 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.
Two pendulums have equal lengths $L$ but different small amplitudes. Which statement about their periods is correct?
Answer: They have the same period because, for small angles, $T$ depends on $L$ and $g$ . 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.
A horizontal mass–spring system ( $m$ , $k$ ) oscillates in SHM. If the system is moved to a planet with different $g$ but kept horizontal, the period
Answer: is unchanged This question examines how gravitational field strength affects different types of oscillators. For a horizontal mass-spring system, the period is $T = 2\pi \sqrt{\frac{m}{k}}$ , which depends only on mass $m$ and spring constant $k$ , not on gravitational field strength $g$ . This is because in horizontal motion, gravity acts perpendicular to the motion and doesn't contribute to the restoring force - only the spring provides the restoring force. The spring's properties ( $k$ ) and the mass's inertia ( $m$ ) remain unchanged on different planets. Choice A incorrectly suggests the period increases with smaller $g$ , confusing horizontal springs with vertical springs or pendulums where $g$ does matter. When analyzing SHM systems, always identify whether gravity contributes to the restoring force - it does for pendulums and vertical springs, but not for horizontal springs.
Two blocks with masses $m$ and $9m$ oscillate on identical springs of constant $k$ with small amplitude. Compared to the $m$ block, the $9m$ block's period is
Answer: $3T$ This question examines how mass affects period in mass-spring oscillators. The period of a mass-spring system is $T = 2\pi \sqrt{\frac{m}{k}}$ , showing that period is proportional to the square root of mass. When comparing blocks of mass m and 9m on identical springs, the period ratio is $T_2 / T_1 = \sqrt{\frac{9m}{k}} / \sqrt{\frac{m}{k}} = \sqrt{9} = 3$ . The heavier block has a period three times longer because its greater inertia requires more time to accelerate and decelerate through each cycle, despite experiencing the same spring forces. Choice C incorrectly suggests nine times the period, failing to recognize the square root relationship between mass and period. When solving for period changes, always check whether the formula involves a square root - this is characteristic of SHM where restoring force is linear in displacement.
A mass $m$ oscillates in SHM on a spring with constant $k$ . The spring is replaced by one with constant $9k$ while keeping the same mass. How does the frequency change?
Answer: The frequency triples. 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.
Two mass–spring systems undergo SHM: System 1 has mass $m$ and spring constant $k$ ; System 2 has mass $4m$ and spring constant $k$ . What is the relationship between their periods, $T_2$ and $T_1$ ?
Answer: $T_2 = 2T_1$ This question assesses understanding of the frequency and period in simple harmonic motion for mass-spring systems. The period T for a mass-spring oscillator is T = 2π √(m/k), so it increases with the square root of mass when k is constant. Frequency f = 1/T decreases with the square root of mass, meaning heavier masses oscillate more slowly. Here, quadrupling the mass from m to 4m doubles the period since √4 = 2, while amplitude is irrelevant as it does not affect T. A common distractor is choice B, which wrongly quadruples the period by linearly scaling with mass instead of its square root. To solve such problems, recall the standard period formulas for SHM systems and identify which parameters are truly independent of the variable changed.
A block of mass $m$ oscillates on a horizontal spring of constant $k$ with amplitude $A$ . If the amplitude is doubled to $2A$ , how does the period change?
Answer: It is unchanged because $T$ depends only on $m$ and $k$ for SHM. 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.
A mass–spring oscillator has mass $m$ and spring constant $k$ . If the mass is replaced with $4m$, what happens to the frequency?
Answer: It decreases by a factor of $2$ because $f\propto 1/\sqrt{m}$ . 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.
A mass $m$ on a spring oscillates with frequency $f$ . If the amplitude is reduced to one-half, what happens to the frequency?
Answer: It is unchanged because $f=\tfrac{1}{2\pi}\sqrt{k/m}$ This question tests whether amplitude affects frequency in SHM. For a mass-spring system, frequency f = (1/2π)√(k/m) depends only on mass and spring constant, not on amplitude. When amplitude is halved, the mass travels half the distance per cycle but also moves proportionally slower at each position, so the time per cycle (and thus frequency) remains unchanged. Choice B incorrectly assumes that less distance means higher frequency, ignoring that maximum speed also decreases proportionally with amplitude. The fundamental principle is that in SHM, frequency and period are determined solely by system parameters (m, k for springs; L, g for pendulums), never by amplitude.
A block oscillates on a spring with period $T$ . The same block is attached to a spring with constant $4k$ instead. What is the new period?
Answer: $\tfrac{T}{2}$ because $T\propto 1/\sqrt{k}$ This question tests understanding of how spring constant affects period. Since period T = 2π√(m/k), when spring constant increases from k to 4k, the new period becomes T' = 2π√(m/4k) = (1/2)·2π√(m/k) = T/2. The quadrupled spring constant provides four times the restoring force for any given displacement, causing the mass to accelerate faster and complete oscillations in less time. Choice C incorrectly assumes period increases with spring stiffness, contradicting the physical intuition that stiffer springs cause faster oscillations. The key insight is that period varies as 1/√k, so multiplying k by 4 divides period by 2.
A simple pendulum of length $L$ oscillates with small amplitude on Earth. The same pendulum is taken to a planet where the gravitational field is $4g$ . How does its period change?
Answer: The period becomes half as large because $T \propto 1/\sqrt{g}$ . This question assesses understanding of the frequency and period in simple harmonic motion for simple pendulums. The period T of a simple pendulum is T = 2π √(L/g), so it depends inversely on the square root of gravitational acceleration g when length L is constant. Frequency f increases with √g, meaning stronger gravity leads to faster oscillations. On a planet with 4g, the period halves since √4 = 2 and T ∝ 1/√g. A common distractor is choice B, which incorrectly quadruples the period by linearly scaling with g instead of using the inverse square root. To solve such problems, recall the standard period formulas for SHM systems and identify which parameters are truly independent of the variable changed.
A simple pendulum of length $L$ swings at small angles with period $T$ . If its length is changed to $\tfrac{L}{9}$ , what is the new period?
Answer: $\tfrac{T}{3}$ because $T\propto \sqrt{L}$ This question tests how pendulum length affects period. The period of a simple pendulum is T = 2π√(L/g), showing that period is proportional to the square root of length. When length changes from L to L/9, the new period becomes T' = 2π√(L/9g) = (1/3)·2π√(L/g) = T/3. A shorter pendulum has a smaller radius of motion and completes swings more quickly, even though the angular acceleration from gravity remains the same. Choice B incorrectly assumes direct proportionality between period and length (T ∝ L) rather than the square root relationship (T ∝ √L). Remember that halving pendulum length doesn't halve the period—it reduces period by a factor of 1/√2.
A mass $m$ oscillates on a spring with constant $k$ and period $T$ . If both $m$ and $k$ are doubled, what happens to the period?
Answer: It is unchanged because $T=2\pi\sqrt{m/k}$ and the ratio $m/k$ is the same. 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.
A mass $m$ oscillates on a spring of constant $k$ with frequency $f$ . If the spring is replaced by one with constant $k/9$ , what is the new frequency?
Answer: $f/3$ , because $f\propto\sqrt{k}$ . This question tests how spring constant affects oscillation frequency. The frequency formula f = (1/2π)√(k/m) shows that frequency is proportional to the square root of spring constant. When k is replaced by k/9, the new frequency becomes f' = (1/2π)√(k/9m) = (1/3) × (1/2π)√(k/m) = f/3. A softer spring (smaller k) provides weaker restoring force, leading to slower oscillations and lower frequency. Choice B incorrectly assumes a linear relationship between k and f, missing the square root dependence. The strategy is to remember that both frequency and period involve square roots of the ratio k/m, reflecting the balance between restoring force and inertia.
Two identical springs (each constant $k$ ) support a mass $m$ . If arranged in parallel, what happens to the oscillation period compared with one spring?
Answer: It decreases by a factor of $\sqrt{2}$ because the effective spring constant is $2k$. 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.
Two mass–spring oscillators have the same mass $m$ , but spring constants $k$ and $4k$. Which has the greater frequency?
Answer: The $4k$ spring, because $f\propto\sqrt{k}$ . This question tests the relationship between spring constant and oscillation frequency. For a mass-spring system, the frequency is f = (1/2π)√(k/m), showing that frequency is proportional to the square root of the spring constant. Since one spring has constant 4k while the other has k, and both have the same mass, the stiffer spring (4k) will have frequency 2 times greater than the softer spring. Choice A incorrectly suggests softer springs oscillate faster, when actually stiffer springs provide stronger restoring forces and thus faster oscillations. The key insight is that frequency increases with spring stiffness because the restoring force F = -kx is stronger for larger k.
A mass $m$ on a spring oscillates with spring constant $k$ . If the mass is replaced by $4m$, what happens to the frequency?
Answer: It becomes half as large because $f\propto 1/\sqrt{m}$ This question examines how mass affects frequency in a mass-spring oscillator. The frequency of a mass-spring system is f = (1/2π)√(k/m), showing that frequency is inversely proportional to the square root of mass. When mass increases from m to 4m, the frequency becomes f' = (1/2π)√(k/4m) = (1/2)(1/2π)√(k/m) = f/2. This happens because a larger mass has more inertia, making it harder to accelerate and thus oscillating more slowly. Choice B incorrectly suggests frequency doubles with mass, confusing the effect of inertia. To solve such problems, use the frequency formula and substitute the new value to find the ratio.
A simple pendulum of length $L$ swings with small angle amplitude. If the bob's mass is increased, how does the period change?
Answer: The period is unchanged because $T=2\pi\sqrt{L/g}$ This question tests understanding of simple pendulum period dependence. 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. This counterintuitive result occurs because while a heavier bob experiences a larger gravitational force, it also has proportionally more inertia to overcome, and these effects exactly cancel out. Choice A incorrectly assumes that more inertia means slower motion, ignoring that the restoring force also increases proportionally. The strategy is to remember that for a simple pendulum, only length and gravity matter for period—mass cancels out completely.

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