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 increases because the weight is smaller.
It decreases because the mass accelerates less.
It is unchanged because $T=2\pi\sqrt{m/k}$ does not involve $g$.
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.
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 halves because the spring is stiffer.
It increases because the amplitude will increase when $k$ increases.
It doubles because the mass doubled.
It is unchanged because $T=2\pi\sqrt{m/k}$ and the ratio $m/k$ is the same.
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 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?
$\tfrac{T}{3}$ because $T\propto 1/\sqrt{k}$.
$3T$ because the block moves three times faster.
$\tfrac{T}{9}$ because $T\propto 1/k$.
$9T$ because a stiffer spring makes the force larger.
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 is unchanged because horizontal mass–spring frequency depends on $m$ and $k$, not $g$.
It increases if $g$ increases because the weight increases.
It changes because frequency equals maximum speed divided by amplitude.
It decreases if $g$ decreases because the amplitude becomes larger.
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.
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 is unchanged because the length is the same.
It increases because $T\propto 1/\sqrt{g}$.
It depends on the bob’s mass, which is unchanged.
It decreases because weaker gravity means less time to fall.
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 small-angle pendulum of length $L$ oscillates with period $T$. If the amplitude is increased but still small enough for SHM, what happens to $T$?
It remains approximately the same because $T$ depends mainly on $L$ and $g$.
It changes because the restoring force depends on amplitude.
It decreases because the maximum speed increases.
It increases because the bob travels farther each cycle.
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.
A mass–spring oscillator has mass $m$ and spring constant $k$. If the mass is replaced with $4m$, what happens to the frequency?
It increases by a factor of $2$ because the maximum speed decreases.
It doubles because the inertia is larger.
It is unchanged because frequency depends on amplitude, not mass.
It decreases by a factor of $2$ because $f\propto 1/\sqrt{m}$.
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.
A simple pendulum of length $L$ swings with small angle amplitude. If its length is increased to $4L$, what happens to its period?
It halves because the restoring force increases with length.
It stays the same because period depends only on mass.
It doubles because $T\propto \sqrt{L}$.
It becomes $4$ times larger because the bob moves through a longer arc.
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.
Two horizontal mass–spring oscillators have the same spring constant $k$. Oscillator X has amplitude $A$ and oscillator Y has amplitude $3A$ (same mass). How do their frequencies compare?
$f_Y = f_X$ because frequency depends only on $m$ and $k$.
$f_Y > f_X$ because the maximum restoring force is larger.
$f_Y = 3f_X$ because the amplitude is three times larger.
$f_Y = \tfrac{1}{3}f_X$ because a larger amplitude makes each cycle longer.
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.
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?
The frequency is unchanged because the amplitude is unchanged.
The frequency decreases by a factor of 3 because the spring is stiffer.
The frequency increases by a factor of 9.
The frequency triples.
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.