Frequency and Period of SHM - AP Physics C Mechanics

Using the given parameters, a simple pendulum of length $L=0.90\,\text{m}$ (small angle) is released; what is the period?
Answer: $T=2\pi\sqrt{\dfrac{L}{g}}\,\text{s}$ 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 undergoing small-angle oscillations, the period T is determined by T = 2π√(L/g), where L is length and g is gravitational acceleration. Choice A is correct because it shows the proper pendulum formula T = 2π√(L/g), which with L = 0.90 m and g = 9.8 m/s² gives T = 2π√(0.90/9.8) = 1.90 s. Choice C is incorrect because it uses the mass-spring formula, which doesn't apply to pendulums - pendulum period is independent of mass. To help students: Distinguish between mass-spring (T = 2π√(m/k)) and pendulum (T = 2π√(L/g)) systems. Emphasize that pendulum period depends only on length and gravity, not on mass or amplitude (for small angles).
Based on the described system, a small-angle pendulum with $L=1.6\,\text{m}$ oscillates; calculate the frequency in Hz using $f=1/T$ .
Answer: $f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}\,\text{Hz}$ 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 is T = 2π√(L/g), so the frequency f = 1/T = 1/(2π√(L/g)) = (1/2π)√(g/L). Choice A is correct because it properly shows f = (1/2π)√(g/L), which with L = 1.6 m and g = 9.8 m/s² gives f = (1/2π)√(9.8/1.6) = 0.394 Hz. Choice B is incorrect because it inverts the g/L ratio, which would give incorrect units and violate the physics - longer pendulums have lower frequencies. To help students: Show the algebraic manipulation from T to f explicitly. Use dimensional analysis to verify that √(g/L) has units of s⁻¹, confirming the frequency formula is correct.
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?
Answer: $T=2\pi\sqrt{\dfrac{m}{k}}$ 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.
Using the given parameters, a $1.2\,\text{kg}$ block on a $k=300\,\text{N/m}$ spring oscillates; how does period change if mass doubles?
Answer: It increases by $\sqrt{2}$ : $T' = \sqrt{2}\,T$ 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), showing that period is proportional to √m. Choice B is correct because when mass doubles from m to 2m, the new period T' = 2π√(2m/k) = √2 × 2π√(m/k) = √2 × T. Choice A is incorrect because it assumes a linear relationship between mass and period, when the actual relationship involves the square root. To help students: Emphasize that T ∝ √m, not T ∝ m, which means doubling mass increases period by √2 ≈ 1.414. Practice problems where students predict how period changes when mass is multiplied by various factors (4, 9, 1/4) to reinforce the square root relationship.
Using the given parameters, a pendulum of length $0.90\,\text{m}$ in $g=9.8\,\text{m/s}^2$ is released from rest at $5^\circ$ ; what is the period of oscillation?
Answer: $T=2\pi\sqrt{\dfrac{L}{g}}$ 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 the pendulum length and g is gravitational acceleration. Choice B is correct because it properly applies the pendulum period formula T = 2π√(L/g), which gives the time for one complete oscillation regardless of the small initial angle. Choice A is incorrect because it inverts the ratio under the square root, giving T = 2π√(g/L), which would have incorrect units and violate the physics of pendulum motion. To help students: Emphasize dimensional analysis - period must have units of time, achieved only with L/g under the square root. Practice deriving this formula from the small-angle approximation to build understanding.
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?
Answer: $f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}$ 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.
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?
Answer: $2\pi\sqrt{\frac{m}{k}}$ 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.
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?
Answer: $T_0\sqrt{\frac{2}{3}}$ 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.
Using the given parameters, a $0.50\,\text{kg}$ block on a $k=200\,\text{N/m}$ spring oscillates; what is the period?
Answer: $T=2\pi\sqrt{\dfrac{m}{k}}\,\text{s}$ 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.
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.
Answer: $f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ 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.
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?
Answer: $2$ 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.
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.
Answer: $T=2\pi\sqrt{\dfrac{m}{k}}$ 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.
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?
Answer: Frequency decreases by $\sqrt{2}$ : $f' = \tfrac{f}{\sqrt{2}}$ . 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.
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?
Answer: It increases by $\sqrt{2}$ : $T' = \sqrt{2}\,T$ . 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).
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?
Answer: $T_2 = T_1$ 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.
Using the given parameters, a pendulum of length $L=2.0\,\text{m}$ swings at small angles; determine the time for one full oscillation.
Answer: $T=2\pi\sqrt{\dfrac{L}{g}}\,\text{s}$ 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.
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?
Answer: The period decreases by a factor of 4 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.
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.
Answer: $f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\,\text{Hz}$ 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.
Based on the described system, a $0.40\,\text{kg}$ mass on a $k=160\,\text{N/m}$ spring oscillates; what is the effect on frequency if $k$ is halved?
Answer: Frequency decreases by $\sqrt{2}$ : $f' = f/\sqrt{2}$ 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), showing that frequency is proportional to √k. Choice C is correct because when k is halved from k to k/2, the new frequency f' = (1/2π)√(k/2m) = (1/√2) × (1/2π)√(k/m) = f/√2. Choice B is incorrect because it assumes a linear inverse relationship between spring constant and frequency, missing the square root dependence. To help students: Highlight that f ∝ √k, so halving k reduces frequency by 1/√2 ≈ 0.707. Create a table showing how frequency changes with different spring constant multipliers to visualize the square root relationship.
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 $\frac{1}{2}MR^2$ )
Answer: $\frac{1}{2\pi}\sqrt{\frac{2g}{3R}}$ 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.

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