Frequency and Period of SHM
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AP Physics 1 › Frequency and Period of SHM
A mass $m$ oscillates on a spring with constant $k$. The oscillator is taken to a location where gravity is smaller, but the surface remains horizontal. What happens to the period?
It changes only if the amplitude changes.
It increases because the weight is smaller.
It is unchanged because $T=2\pi\sqrt{m/k}$ does not involve $g$.
It decreases because the mass accelerates less.
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.
A simple pendulum of length $L$ oscillates with small angle amplitude. A second pendulum has length $4L$ with the same bob mass. How does the second pendulum’s frequency compare to the first?
It is four times as large because the length is four times as large.
It is half as large because frequency is inversely proportional to $\sqrt{L}$.
It is unchanged because frequency depends only on the bob’s mass.
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.
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?
It increases because the amplitude will increase when $k$ increases.
It is unchanged because $T=2\pi\sqrt{m/k}$ and the ratio $m/k$ is the same.
It halves because the spring is stiffer.
It doubles because the mass doubled.
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.
A simple pendulum of length $L$ and bob mass $m$ oscillates with small amplitude. If the bob is replaced by one of mass $2m$, how does the period change?
It increases by a factor of $\sqrt{2}$ because heavier bobs swing more slowly.
It doubles because the mass doubled.
It decreases by a factor of $\sqrt{2}$ because gravity pulls harder.
It is unchanged because for small angles $T$ depends on $L$ and $g$ only.
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.
A block oscillates on a spring with period $T$. If the spring constant is increased to $9k$ while mass stays $m$, what is the new period?
$9T$ because a stiffer spring makes the force larger.
$\tfrac{T}{9}$ because $T\propto 1/k$.
$\tfrac{T}{3}$ because $T\propto 1/\sqrt{k}$.
$3T$ because the block moves three times faster.
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.
A mass $m$ oscillates on a spring on Earth with angular frequency $\omega$. The same setup is taken to a planet with different $g$ but used horizontally. How does the frequency change?
It changes because frequency equals maximum speed divided by amplitude.
It is unchanged because horizontal mass–spring frequency depends on $m$ and $k$, not $g$.
It decreases if $g$ decreases because the amplitude becomes larger.
It increases if $g$ increases because the weight increases.
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.
Two identical springs (each constant $k$) support a mass $m$. If arranged in parallel, what happens to the oscillation period compared with one spring?
It decreases by a factor of 2 because the amplitude is shared by two springs.
It increases by a factor of $\sqrt{2}$ because there are two springs.
It decreases by a factor of $\sqrt{2}$ because the effective spring constant is $2k$.
It is unchanged because the mass is the same.
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.
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?
It increases because $T\propto 1/\sqrt{g}$.
It decreases because weaker gravity means less time to fall.
It depends on the bob’s mass, which is unchanged.
It is unchanged because the length is the same.
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.
A block–spring system oscillates with maximum speed $v_{\max}$ at equilibrium. If the amplitude is increased while $m$ and $k$ stay the same, what happens to the frequency?
It decreases because the block travels farther each cycle.
It increases because the block’s maximum speed increases.
It becomes proportional to amplitude because the restoring force is larger.
It is unchanged because frequency is set by $m$ and $k$.
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.
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?
It halves because the restoring force is larger at larger displacement.
It doubles because the block travels twice as far each cycle.
It increases because frequency depends on amplitude.
It is unchanged because $T$ depends only on $m$ and $k$ for SHM.
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.