AP Physics C Mechanics Quiz: Frequency And Period Of Shm

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Frequency And Period Of ShmQuestion 1 of 20

Using the given parameters, a horizontal mass-spring oscillator has $m=0.40\,\text{kg}$ , $k=160\,\text{N/m}$ , amplitude $0.05\,\text{m}$ , and $v(0)=0$ at $x(0)=+A$ ; how does the period change if the mass is doubled?

It doubles:

T' = 2T

It halves:

T' = \tfrac{1}{2}T

It increases by

\sqrt{2}

T' = \sqrt{2}\,T

It is unchanged:

T' = T

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Frequency And Period Of Shm

Practice Frequency And Period Of Shm in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 C Mechanics.

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.

All questions

Question 1

It doubles: $T' = 2T$ .
It halves: $T' = \tfrac{1}{2}T$ .
It increases by $\sqrt{2}$ : $T' = \sqrt{2}\,T$ . (correct answer)
It is unchanged: $T' = T$ .

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, the period T is determined by T = 2π√(m/k), where m is mass and k is the spring constant. Choice C is correct because when mass doubles (m' = 2m), the new period becomes T' = 2π√(2m/k) = √2 × 2π√(m/k) = √2 × T, showing that period increases by a factor of √2. Choice A is incorrect because it assumes a linear relationship between mass and period, when the actual relationship involves a square root. To help students: Emphasize that period depends on √m, not m directly. Practice problems where system parameters change to reinforce how T scales with √(m/k).

Question 2

Based on the described system, a $0.60\,\text{kg}$ block attached to a spring $k=240\,\text{N/m}$ oscillates with amplitude $0.08\,\text{m}$ ; what is the effect on frequency if the spring constant is halved?

Frequency doubles: $f' = 2f$ .
Frequency decreases by $\sqrt{2}$ : $f' = \tfrac{f}{\sqrt{2}}$ . (correct answer)
Frequency halves: $f' = \tfrac{1}{2}f$ .
Frequency is unchanged: $f' = f$ .

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, the frequency f is given by f = (1/2π)√(k/m), where k is the spring constant and m is mass. Choice B is correct because when the spring constant is halved (k' = k/2), the new frequency becomes f' = (1/2π)√(k/2m) = (1/√2) × (1/2π)√(k/m) = f/√2, showing frequency decreases by a factor of √2. Choice C is incorrect because it assumes a linear relationship between spring constant and frequency, missing the square root dependence. To help students: Emphasize that frequency depends on √k, not k directly. Use dimensional analysis to verify that √(k/m) has units of 1/time, appropriate for frequency.

Question 3

Using the given parameters, a $0.50\,\text{kg}$ block on a frictionless surface is attached to a spring $k=200\,\text{N/m}$ , amplitude $0.10\,\text{m}$ , released from rest at maximum stretch; what is the period of oscillation?

$T=2\pi\sqrt{\dfrac{k}{m}}$
$T=2\pi\sqrt{\dfrac{m}{k}}$ (correct answer)
$T=\dfrac{1}{2\pi}\sqrt{\dfrac{m}{k}}$
$T=2\pi\sqrt{\dfrac{m}{g}}$

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, the period T is determined by T = 2π√(m/k), where m is mass and k is the spring constant. Choice B is correct because it accurately applies the formula T = 2π√(m/k) using the given values, demonstrating an understanding that period depends on the square root of mass over spring constant. Choice A is incorrect because it inverts the ratio, suggesting T = 2π√(k/m), which would give incorrect units and violate the physics of oscillation. To help students: Emphasize dimensional analysis - period must have units of time, which only occurs when m/k is under the square root. Practice deriving this formula from F = -kx and Newton's second law to build intuition.

Question 4

Based on the described system, a vertical spring-mass oscillator has $m=0.50\,\text{kg}$ , $k=80\,\text{N/m}$ , and amplitude $0.03\,\text{m}$ about equilibrium; calculate the frequency of the given system in Hz.

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ (correct answer)
$f=2\pi\sqrt{\dfrac{m}{k}}$
$f=\dfrac{1}{2\pi}\sqrt{\dfrac{m}{k}}$
$f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}$

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, whether vertical or horizontal, the frequency f is given by f = (1/2π)√(k/m), where k is the spring constant and m is mass. Choice A is correct because it properly applies the frequency formula f = (1/2π)√(k/m), which gives oscillations per second (Hz) for the mass-spring system. Choice C is incorrect because it has m/k under the square root instead of k/m, which would give incorrect units and violate the physics - stiffer springs (larger k) should produce higher frequencies. To help students: Emphasize that vertical and horizontal springs use the same frequency formula. Practice unit analysis to verify that √(k/m) has units of 1/time.

Question 5

Using the given parameters, a horizontal spring-mass system has $m=0.80\,\text{kg}$ , $k=50\,\text{N/m}$ , amplitude $0.12\,\text{m}$ , and initial velocity $v(0)=+0.30\,\text{m/s}$ at $x(0)=0$ ; determine the time it takes to complete one full oscillation.

$T=2\pi\sqrt{\dfrac{m}{k}}$ (correct answer)
$T=\pi\sqrt{\dfrac{m}{k}}$
$T=2\pi\sqrt{\dfrac{k}{m}}$
$T=2\pi\sqrt{\dfrac{L}{g}}$

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, the period T is determined by T = 2π√(m/k), where m is mass and k is the spring constant. Choice A is correct because the period formula T = 2π√(m/k) applies regardless of initial conditions - whether starting from rest at maximum displacement or with initial velocity at equilibrium, the time for one complete cycle remains the same. Choice B is incorrect because it omits the factor of 2 in the formula, giving T = π√(m/k), which would yield half the correct period. To help students: Emphasize that period is a property of the system (m and k), not dependent on initial conditions or amplitude. Practice problems with various starting conditions to reinforce this concept.

Question 6

Using the given parameters, a $0.30\,\text{kg}$ block on a $k=120\,\text{N/m}$ spring has amplitude $0.10\,\text{m}$ ; explain why period remains constant.

Because $T=2\pi\sqrt{m/k}$ is independent of amplitude for ideal SHM (correct answer)
Because $T$ increases linearly with amplitude: $T\propto A$ for springs
Because $T$ depends on maximum speed: larger amplitude always shortens $T$
Because $T=2\pi\sqrt{L/g}$ and amplitude cancels for a spring

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For an ideal mass-spring system, the period T = 2π√(m/k) depends only on mass and spring constant, not on amplitude. Choice A is correct because it correctly states that period is independent of amplitude in ideal SHM, a fundamental property that distinguishes linear from nonlinear oscillators. Choice B is incorrect because it claims period increases with amplitude, which would violate the isochronous nature of SHM. To help students: Demonstrate experimentally that different amplitudes yield the same period. Explain that this independence breaks down for large amplitudes in real systems (anharmonic effects) but holds for ideal SHM.

Question 7

Two identical springs with spring constant $k$ are connected in series, and a mass $m$ is attached to the free end. The system undergoes simple harmonic motion. If one spring is removed and replaced with a rigid connection, how does the frequency change?

The frequency increases by a factor of $\sqrt{2}$ (correct answer)
The frequency increases by a factor of $2$
The frequency decreases by a factor of $\sqrt{2}$
The frequency remains unchanged since the mass is the same

Explanation: For springs in series, the effective spring constant is

\frac{1}{k_{eff}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}

, so

k_{eff} = \frac{k}{2}

. The initial frequency is

f_1 = \frac{1}{2\pi}\sqrt{\frac{k/2}{m}}

. After removing one spring, the effective spring constant becomes

k

, so

f_2 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

. Therefore,

\frac{f_2}{f_1} = \sqrt{\frac{k}{k/2}} = \sqrt{2}

. Choice B forgets the square root. Choice C inverts the relationship. Choice D ignores the change in effective spring constant.

Question 8

A simple pendulum of length $L$ has period $T$ on Earth. The same pendulum is taken to a planet where the gravitational acceleration is $g' = 4g$ and oscillates with the same amplitude. What is the ratio of the frequency on the planet to the frequency on Earth?

$\frac{1}{4}$
$\frac{1}{2}$
$2$ (correct answer)
$4$

Explanation: For a simple pendulum,

T = 2\pi\sqrt{\frac{L}{g}}

, so

f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}

. On Earth:

f_E = \frac{1}{2\pi}\sqrt{\frac{g}{L}}

. On the planet:

f_P = \frac{1}{2\pi}\sqrt{\frac{4g}{L}} = 2 \cdot \frac{1}{2\pi}\sqrt{\frac{g}{L}} = 2f_E

. Therefore,

\frac{f_P}{f_E} = 2

. Choice A gives the period ratio instead of frequency ratio. Choice B takes the square root incorrectly. Choice D uses the gravitational acceleration ratio directly.

Question 9

Two identical masses $m$ are connected by a spring with spring constant $k$ and can slide without friction on a horizontal surface. When the system undergoes oscillations where the masses move in opposite directions with equal amplitudes, what is the frequency of this normal mode?

$\frac{1}{2\pi}\sqrt{\frac{k}{m}}$
$\frac{1}{2\pi}\sqrt{\frac{2k}{m}}$ (correct answer)
$\frac{1}{2\pi}\sqrt{\frac{k}{2m}}$
$\frac{1}{\pi}\sqrt{\frac{k}{m}}$

Explanation: In the antisymmetric mode where masses move in opposite directions, the center of mass remains fixed. From each mass's perspective, the other mass acts as a fixed wall, doubling the effective spring constant. The equation of motion for each mass is

m\ddot{x} = -kx - k(x-0) = -2kx

, giving

\omega = \sqrt{\frac{2k}{m}}

and

f = \frac{1}{2\pi}\sqrt{\frac{2k}{m}}

. Choice A gives the frequency for a single mass-spring system. Choice C incorrectly doubles the mass instead of the spring constant. Choice D has the wrong factor in the denominator.

Question 10

A mass $m$ is attached to a spring with spring constant $k$ and undergoes simple harmonic motion. At the moment when the displacement is $x = \frac{A}{2}$ (where $A$ is the amplitude), the speed is $v = \frac{A\sqrt{3}}{2}\omega$ , where $\omega$ is the angular frequency. What is the period of this motion?

The given conditions are physically impossible
$\pi\sqrt{\frac{m}{k}}$
$4\pi\sqrt{\frac{m}{k}}$
$2\pi\sqrt{\frac{m}{k}}$ (correct answer)

Explanation: When you encounter simple harmonic motion problems, always start with the fundamental energy relationship. The total mechanical energy in SHM remains constant and equals the maximum potential energy:

E = \frac{1}{2}kA^2

. At any position, this energy splits between kinetic and potential:

E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2

. Let's check if the given conditions are physically consistent by substituting the values. With

x = \frac{A}{2}

and

v = \frac{A\sqrt{3}}{2}\omega

\frac{1}{2}kA^2 = \frac{1}{2}m\left(\frac{A\sqrt{3}}{2}\omega\right)^2 + \frac{1}{2}k\left(\frac{A}{2}\right)^2

\frac{1}{2}kA^2 = \frac{1}{2}m \cdot \frac{3A^2\omega^2}{4} + \frac{1}{2}k \cdot \frac{A^2}{4}

\frac{1}{2}kA^2 = \frac{3mA^2\omega^2}{8} + \frac{kA^2}{8}

Dividing by

\frac{A^2}{2}

k = \frac{3m\omega^2}{4} + \frac{k}{4}

Solving:

\frac{3k}{4} = \frac{3m\omega^2}{4}

, so

k = m\omega^2

This gives us

\omega = \sqrt{\frac{k}{m}}

, which is exactly the standard formula for SHM angular frequency. Therefore, the period is

T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}

. Answer choice A is wrong because the conditions are physically possible, as we just verified. Choice B gives half the correct period, while choice C gives twice the correct period—both likely stem from errors in the

2\pi

factor. Study tip: Always verify that given SHM conditions satisfy energy conservation before solving. This catches impossible scenarios and confirms your approach is correct.

Question 11

A block of mass $m$ slides without friction on a horizontal surface and is connected to a spring with spring constant $k$ . At $t = 0$ , the block is at position $x = A$ and has velocity $v = -A\omega$ , where $\omega = \sqrt{\frac{k}{m}}$ . What is the period of the resulting motion?

$\frac{2\pi}{\omega}$ (correct answer)
$\frac{\pi}{\omega}$
$\frac{4\pi}{\omega}$
$\frac{\pi}{2\omega}$

Explanation: The given initial conditions describe simple harmonic motion with the block starting at maximum displacement with maximum velocity toward equilibrium. This is standard SHM with angular frequency

\omega = \sqrt{\frac{k}{m}}

and period

T = \frac{2\pi}{\omega}

. The specific initial conditions don't change the period - they only determine the phase. Choice B gives half the correct period. Choice C gives twice the correct period. Choice D gives one-fourth the correct period.

Question 12

A spring-mass system has a natural frequency $f_0$ . When the system is driven by an external sinusoidal force at frequency $f$ , steady-state oscillations occur. If the driving frequency is suddenly changed from $f = 0.5f_0$ to $f = 2f_0$ , how does the period of the steady-state oscillations change?

The period increases by a factor of 4
The period decreases by a factor of 4 (correct answer)
The period increases by a factor of 2
The period remains equal to $\frac{1}{f_0}$

Explanation: In steady-state driven oscillations, the system oscillates at the driving frequency, not the natural frequency. Initially,

T_1 = \frac{1}{f} = \frac{1}{0.5f_0} = \frac{2}{f_0}

. After the change,

T_2 = \frac{1}{2f_0}

. Therefore,

\frac{T_2}{T_1} = \frac{1/2f_0}{2/f_0} = \frac{1}{4}

, so the period decreases by a factor of 4. Choice A inverts the relationship. Choice C uses the frequency ratio instead of its reciprocal. Choice D incorrectly assumes oscillation at natural frequency.

Question 13

A spring-mass system undergoes simple harmonic motion with amplitude $A$ and period $T_0$ . If the mass is doubled and the spring constant is tripled, while the amplitude remains unchanged, what is the new period?

$T_0\sqrt{\frac{2}{3}}$ (correct answer)
$T_0\sqrt{\frac{3}{2}}$
$\frac{T_0}{\sqrt{6}}$
$T_0\sqrt{6}$

Explanation: The period of a spring-mass system is

T = 2\pi\sqrt{\frac{m}{k}}

. Initially,

T_0 = 2\pi\sqrt{\frac{m}{k}}

. When mass becomes

2m

and spring constant becomes

3k

, the new period is

T_{new} = 2\pi\sqrt{\frac{2m}{3k}} = 2\pi\sqrt{\frac{2}{3}}\sqrt{\frac{m}{k}} = T_0\sqrt{\frac{2}{3}}

. Choice B inverts the fraction. Choice C incorrectly multiplies the factors instead of taking the square root. Choice D assumes both factors add under the square root.

Question 14

A uniform disk of radius $R$ and mass $M$ is suspended from a point on its rim and oscillates as a physical pendulum. What is the frequency of small oscillations? (The moment of inertia of a disk about its center is $\frac{1}{2}MR^2$ )

$\frac{1}{2\pi}\sqrt{\frac{3g}{2R}}$
$\frac{1}{2\pi}\sqrt{\frac{g}{R}}$
$\frac{1}{2\pi}\sqrt{\frac{2g}{3R}}$ (correct answer)
$\frac{1}{2\pi}\sqrt{\frac{g}{2R}}$

Explanation: When you encounter a physical pendulum problem, you need to apply the general formula for the period of a physical pendulum:

T = 2\pi\sqrt{\frac{I}{mgd}}

, where

I

is the moment of inertia about the pivot point,

m

is mass,

g

is gravitational acceleration, and

d

is the distance from the pivot to the center of mass. For this disk suspended from its rim, the distance from the pivot to the center of mass is

d = R

. However, you can't use the given moment of inertia

\frac{1}{2}MR^2

directly because that's about the center of the disk, not the pivot point. You must use the parallel axis theorem:

I_{pivot} = I_{center} + Md^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2

. Substituting into the period formula:

T = 2\pi\sqrt{\frac{\frac{3}{2}MR^2}{MgR}} = 2\pi\sqrt{\frac{3R}{2g}}

Since frequency

f = \frac{1}{T}

, we get

f = \frac{1}{2\pi}\sqrt{\frac{2g}{3R}}

, which is answer C. Answer A gives

\sqrt{\frac{3g}{2R}}

, which incorrectly flips the fraction in the square root. Answer B gives

\sqrt{\frac{g}{R}}

, the result for a simple pendulum of length

R

, ignoring the rotational inertia. Answer D gives

\sqrt{\frac{g}{2R}}

, which would result from using only the center moment of inertia without applying the parallel axis theorem. Study tip: Always check whether the given moment of inertia is about the correct axis—physical pendulum problems frequently require the parallel axis theorem.

Question 15

A mass $m$ hangs from a spring and oscillates vertically with period $T_1$ . The same mass is then placed on a horizontal surface and connected to an identical spring. If the coefficient of kinetic friction is $\mu$ and the amplitude is small enough that the mass doesn't stick, what is the period $T_2$ of horizontal oscillations?

$T_2 = \frac{T_1}{\sqrt{1 + \mu}}$
$T_2 = T_1\sqrt{1 + \frac{\mu g}{k A}}$
$T_2 = T_1\sqrt{1 - \frac{\mu mg}{k}}$
$T_2 = T_1$ (correct answer)

Explanation: When analyzing oscillating systems, the key insight is understanding what factors actually affect the period of oscillation. For a mass-spring system, the period depends only on the inertial properties (mass) and the restoring force mechanism (spring constant). The period of oscillation for any mass-spring system is given by

T = 2\pi\sqrt{\frac{m}{k}}

. In the vertical case, gravity shifts the equilibrium position but doesn't change the dynamics of oscillation around that equilibrium. The spring constant

k

and mass

m

remain the same when you move to the horizontal setup. Friction is a non-conservative force that removes energy from the system, causing the amplitude to decrease over time. However, friction does not change the period of oscillation. The mass still accelerates and decelerates according to the net restoring force from the spring, and the time it takes to complete each cycle remains unchanged. The oscillations simply become smaller with each cycle until they eventually stop. Choice A incorrectly suggests friction increases the effective "stiffness" of the system. Choice B wrongly implies that friction creates an amplitude-dependent period, which would make the motion non-harmonic. Choice C incorrectly treats friction as if it reduces the effective spring constant. All of these choices reflect the common misconception that friction affects the timing of oscillations rather than just their energy. Study tip: Remember that period depends only on

m

and

k

for harmonic motion. Friction affects amplitude and energy, but not the fundamental timing of oscillations.

Question 16

Using the given parameters, a pendulum of length $L=2.0\,\text{m}$ swings at small angles; determine the time for one full oscillation.

$T=\pi\sqrt{\dfrac{L}{g}}\,\text{s}$
$T=2\pi\sqrt{\dfrac{L}{g}}\,\text{s}$ (correct answer)
$T=2\pi\sqrt{\dfrac{g}{L}}\,\text{s}$
$T=2\pi\sqrt{\dfrac{m}{k}}\,\text{s}$

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a simple pendulum at small angles, the period T is determined by T = 2π√(L/g), where L is the pendulum length and g is gravitational acceleration. Choice B is correct because it shows the standard pendulum formula T = 2π√(L/g), which with L = 2.0 m and g = 9.8 m/s² gives T = 2π√(2.0/9.8) = 2.84 s. Choice A is incorrect because it's missing the factor of 2 in front of π, which would give half the actual period. To help students: Derive the pendulum formula from the restoring torque τ = -mgL sin θ ≈ -mgLθ for small angles. Practice calculating periods for different pendulum lengths to build intuition about the √L dependence.

Question 17

Based on the described system, a $0.80\,\text{kg}$ mass on a $k=50\,\text{N/m}$ spring oscillates; calculate the frequency in Hz.

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\,\text{Hz}$ (correct answer)
$f=2\pi\sqrt{\dfrac{k}{m}}\,\text{Hz}$
$f=\dfrac{1}{2\pi}\sqrt{\dfrac{m}{k}}\,\text{Hz}$
$f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}\,\text{Hz}$

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, the frequency f is determined by f = (1/2π)√(k/m), where k is the spring constant and m is mass. Choice A is correct because it properly shows f = (1/2π)√(k/m), which with k = 50 N/m and m = 0.80 kg gives f = (1/2π)√(50/0.80) = 1.26 Hz. Choice C is incorrect because it inverts the k/m ratio, resulting in incorrect units and a physically meaningless result. To help students: Connect frequency to the period formula by showing f = 1/T = 1/(2π√(m/k)) = (1/2π)√(k/m). Use unit analysis to verify that √(k/m) has units of s⁻¹, making the frequency have proper Hz units.

Question 18

Based on the described system, a lightly damped oscillator with $m=1.0\,\text{kg}$ and $k=100\,\text{N/m}$ has small damping; what best describes its period?

Approximately $T\approx 2\pi\sqrt{m/k}$ ; damping mainly reduces amplitude (correct answer)
Exactly $T=2\pi\sqrt{m/k}$ is impossible; damping makes $T$ undefined
Period decreases to $T=\dfrac{1}{2\pi}\sqrt{m/k}$ due to damping
Period becomes $T=2\pi\sqrt{k/m}$ because damping reverses the ratio

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a lightly damped oscillator, the period remains approximately T ≈ 2π√(m/k), with damping primarily affecting amplitude decay rather than period. Choice A is correct because light damping causes exponential amplitude decay while leaving the period nearly unchanged - the actual period is slightly longer but the difference is negligible for light damping. Choice B is incorrect because damping doesn't make period undefined; the system still oscillates with a well-defined period. To help students: Introduce the damped oscillator equation and show that for light damping (ζ << 1), the period T ≈ T₀ = 2π√(m/k). Use energy arguments to explain why damping reduces amplitude but minimally affects frequency.

Question 19

Using the given parameters, a $0.50\,\text{kg}$ block on a $k=200\,\text{N/m}$ spring oscillates; what is the period?

$T=2\pi\sqrt{\dfrac{k}{m}}\,\text{s}$
$T=2\pi\sqrt{\dfrac{m}{k}}\,\text{s}$ (correct answer)
$T=\dfrac{1}{2\pi}\sqrt{\dfrac{m}{k}}\,\text{s}$
$T=2\pi\sqrt{\dfrac{L}{g}}\,\text{s}$

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, the period T is determined by T = 2π√(m/k), where m is mass and k is the spring constant. Choice B is correct because it shows the proper formula T = 2π√(m/k), which when calculated with m = 0.50 kg and k = 200 N/m gives T = 2π√(0.50/200) = 0.314 s. Choice A is incorrect because it inverts the mass and spring constant ratio, which would give units of s⁻¹ rather than seconds. To help students: Emphasize dimensional analysis to verify that √(m/k) has units of time. Practice deriving the period formula from F = -kx and Newton's second law to build deeper understanding.

Question 20

Based on the described system, a simple pendulum of length $0.80\,\text{m}$ is released from rest at $6^\circ$ ; using $T=2\pi\sqrt{L/g}$ , what is the frequency in Hz?

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}$ (correct answer)
$f=2\pi\sqrt{\dfrac{L}{g}}$
$f=\dfrac{1}{2\pi}\sqrt{\dfrac{L}{g}}$
$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a simple pendulum, the period T is given by T = 2π√(L/g), where L is length and g is gravitational acceleration. Choice A is correct because it properly converts period to frequency using f = 1/T = 1/(2π√(L/g)) = (1/2π)√(g/L), showing understanding of the inverse relationship. Choice B is incorrect because it represents the period formula, not frequency, demonstrating confusion between these reciprocal quantities. To help students: Emphasize the distinction between period (time per cycle) and frequency (cycles per time). Practice converting between T and f using f = 1/T consistently across different SHM systems.