Representing and Analyzing SHM - AP Physics C Mechanics

A mass $m$ is attached to two identical springs, each with spring constant $k$ , arranged in parallel (both attached to the mass and to fixed supports). The system oscillates with simple harmonic motion. If one spring suddenly breaks when the mass is at maximum displacement $+A$ , what will be the new amplitude of oscillation?
Answer: $\sqrt{2}A$ Initially, the effective spring constant is $k_{eff} = 2k$ (parallel springs). The total energy is $E_1 = \frac{1}{2}(2k)A^2 = kA^2$ . When one spring breaks at maximum displacement, the mass momentarily has zero velocity, so all energy is potential: $E_1 = kA^2$ . With only one spring remaining, $k_{new} = k$ , and this energy becomes $E_2 = \frac{1}{2}kA_{new}^2$ . By energy conservation: $kA^2 = \frac{1}{2}kA_{new}^2$ , so $A_{new} = \sqrt{2}A$ . Choice A would result from incorrectly assuming $E \propto A$ . Choice B ignores the change in spring constant. Choice D assumes the displacement doubles, which violates energy conservation.
A 0.20 kg mass is attached to a horizontal spring with $k=50\ \text{N/m}$ on a frictionless surface, with equilibrium at $x=0$ . The mass is displaced to $x=+0.060\ \text{m}$ and released from rest at $t=0$ , so $A=0.060\ \text{m}$ and $\phi=0$ . The SHM model is $x(t)=A\cos(\omega t+\phi)$ , where $\omega=\sqrt{k/m}$ in rad/s. The speed is $v(t)=-A\omega\sin(\omega t+\phi)$ and the maximum speed occurs at $x=0$ . The restoring force is $F_s=-kx$ and mechanical energy is conserved. What is the maximum speed at equilibrium?
Answer: $v_{\max}=0.95\ \text{m/s}$ This question tests the ability to represent and analyze simple harmonic motion (SHM) in the context of AP Physics C: Mechanics. SHM is characterized by periodic motion where the restoring force is proportional to displacement, commonly modeled by mass-spring systems or pendulums. In this scenario, a 0.20 kg mass on a spring with k=50 N/m is released from rest at amplitude A=0.060 m, and we need to find the maximum speed. Choice A is correct because vmax=Aω where ω=√(k/m)=√(50/0.20)=15.81 rad/s, giving vmax=0.060×15.81≈0.95 m/s. Choice D is incorrect due to a common error where students might divide by 10 instead of multiplying, possibly confusing units or misapplying the formula. To help students: Emphasize that maximum speed occurs at equilibrium where all energy is kinetic, and vmax=Aω is a fundamental SHM relationship. Practice problems should include energy conservation checks to verify that ½mvmax²=½kA².
A frictionless mass–spring oscillator has $m=0.10\ \text{kg}$ , $k=40\ \text{N/m}$ , and equilibrium at $x=0$ . It is released from rest at $x=+0.050\ \text{m}$ at $t=0$ , so $A=0.050\ \text{m}$ and $\phi=0$ . The motion is $x(t)=A\cos(\omega t)$ with $\omega=\sqrt{k/m}$ and period $T=2\pi/\omega$ . The object reaches equilibrium after one quarter period. All quantities use SI units and amplitude remains constant. How long after release does it first reach $x=0$ ?
Answer: $t=0.079\ \text{s}$ This question tests the ability to represent and analyze simple harmonic motion (SHM) in the context of AP Physics C: Mechanics. SHM is characterized by periodic motion where the restoring force is proportional to displacement, commonly modeled by mass-spring systems or pendulums. In this scenario, a 0.10 kg mass on a spring with k=40 N/m is released from rest at maximum displacement, reaching equilibrium after one quarter period. Choice B is correct because ω=√(k/m)=√(40/0.10)=20 rad/s, giving period T=2π/ω=π/10≈0.314 s, so t=T/4≈0.079 s. Choice C is incorrect due to a common error where students might calculate T/2 instead of T/4, thinking the mass needs half a period to reach equilibrium from the extreme position. To help students: Emphasize that SHM motion from extreme position to equilibrium takes exactly one quarter period. Practice problems should include finding times to reach various positions to reinforce the periodic nature of the motion.
A horizontal mass–spring system has $m=0.25\ \text{kg}$ , $k=100\ \text{N/m}$ , and equilibrium at $x=0$ . The mass is released from rest at $x=+0.040\ \text{m}$ at $t=0$ , so $A=0.040\ \text{m}$ and $\phi=0$ . The SHM equation is $x(t)=A\cos(\omega t+\phi)$ with $\omega=\sqrt{k/m}$ . The period is $T=2\pi/\omega$ and the frequency is $f=1/T$ . The phase affects where the motion starts but not $f$ . Assume no friction and constant amplitude. Determine the frequency $f$ of oscillation.
Answer: $f=3.2\ \text{Hz}$ This question tests the ability to represent and analyze simple harmonic motion (SHM) in the context of AP Physics C: Mechanics. SHM is characterized by periodic motion where the restoring force is proportional to displacement, commonly modeled by mass-spring systems or pendulums. In this scenario, a 0.25 kg mass on a spring with k=100 N/m oscillates with angular frequency ω=√(k/m)=√(100/0.25)=20 rad/s. Choice B is correct because the frequency f=ω/(2π)=20/(2π)≈3.18 Hz, which rounds to 3.2 Hz. Choice C is incorrect due to a common error where students might calculate f=1/(2π) without including the angular frequency, or confuse period with frequency. To help students: Emphasize the distinction between angular frequency ω (in rad/s) and frequency f (in Hz), connected by f=ω/(2π). Practice problems should include converting between period, frequency, and angular frequency to build fluency with these related quantities.
Two pendulums of equal length $L$ are released simultaneously from small angular displacements. Pendulum A has mass $m_A$ concentrated at the bob, while Pendulum B has the same total mass $m_A$ distributed as a uniform rod of length $L$ (no additional bob). What is the ratio of their periods $T_A/T_B$ ?
Answer: $\sqrt{\frac{2}{3}}$ For small oscillations, $T = 2\pi\sqrt{\frac{I}{mgd}}$ where $I$ is moment of inertia about pivot, $m$ is mass, and $d$ is distance from pivot to center of mass. Pendulum A (point mass): $I_A = m_A L^2$ , $d_A = L$ , so $T_A = 2\pi\sqrt{\frac{m_A L^2}{m_A g L}} = 2\pi\sqrt{\frac{L}{g}}$ . Pendulum B (uniform rod): $I_B = \frac{1}{3}m_A L^2$ , $d_B = \frac{L}{2}$ , so $T_B = 2\pi\sqrt{\frac{m_A L^2/3}{m_A g L/2}} = 2\pi\sqrt{\frac{2L}{3g}}$ . Therefore $\frac{T_A}{T_B} = \sqrt{\frac{L/g}{2L/3g}} = \sqrt{\frac{3}{2}}$ . But we want $T_A/T_B$ , so $\frac{T_A}{T_B} = \frac{2\pi\sqrt{L/g}}{2\pi\sqrt{2L/3g}} = \sqrt{\frac{3}{2}} = \sqrt{\frac{2}{3}}$ .
A particle undergoes simple harmonic motion with position $x(t) = 0.25\cos(4t + \pi/4)$ meters. At what time $t > 0$ does the particle first reach maximum kinetic energy?
Answer: $\frac{\pi}{16}\text{ s}$ Maximum kinetic energy occurs when the particle passes through equilibrium ( $x = 0$ ) with maximum speed. Setting $x(t) = 0.25\cos(4t + \pi/4) = 0$ , we need $4t + \pi/4 = \pi/2 + n\pi$ where $n$ is an integer. For the first occurrence ( $n = 0$ ): $4t = \pi/2 - \pi/4 = \pi/4$ , so $t = \pi/16\text{ s}$ . Choice B gives $t = \pi/8$ , which would require $4t + \pi/4 = \pi/2 + \pi = 3\pi/2$ . Choice C corresponds to the second zero crossing, and Choice D corresponds to a different phase relationship.
A mass–spring oscillator has $m=0.50\ \text{kg}$ , $k=80\ \text{N/m}$ , and equilibrium at $x=0$ on a frictionless surface. It is displaced to $A=0.10\ \text{m}$ and released from rest, so $x(t)=A\cos(\omega t)$ with $\omega=\sqrt{k/m}$ . The total energy is constant and given by $E=\tfrac12 kA^2$ in joules. At equilibrium the kinetic energy equals $E$ and the spring potential energy is zero. Use SI units throughout and ignore any damping. What is the total mechanical energy $E$ ?
Answer: $E=0.40\ \text{J}$ This question tests the ability to represent and analyze simple harmonic motion (SHM) in the context of AP Physics C: Mechanics. SHM is characterized by periodic motion where the restoring force is proportional to displacement, commonly modeled by mass-spring systems or pendulums. In this scenario, a 0.50 kg mass on a spring with k=80 N/m oscillates with amplitude A=0.10 m, and we need to find the total mechanical energy E=½kA². Choice A is correct because E=½×80×(0.10)²=½×80×0.01=0.40 J. Choice C is incorrect due to a common error where students might forget the factor of ½ in the energy formula, calculating E=kA²/10 or making a decimal place error. To help students: Emphasize that total energy in SHM equals maximum potential energy (at amplitude) or maximum kinetic energy (at equilibrium). Practice problems should include energy calculations at various positions to reinforce conservation of mechanical energy.
A 0.60 kg cart is attached to a spring with $k=240\ \text{N/m}$ and oscillates on a frictionless track about $x=0$ . It is pulled to $x=+0.070\ \text{m}$ and released from rest at $t=0$ , so $A=0.070\ \text{m}$ and $\phi=0$ . The motion is $x(t)=A\cos(\omega t+\phi)$ with $\omega=\sqrt{k/m}$ . The acceleration is $a(t)=-\omega^2 x(t)$ and $a_{\max}=\omega^2 A$ . All quantities are in SI units and amplitude remains constant. Find the maximum acceleration magnitude $a_{\max}$ .
Answer: $a_{\max}=28\ \text{m/s}^2$ This question tests the ability to represent and analyze simple harmonic motion (SHM) in the context of AP Physics C: Mechanics. SHM is characterized by periodic motion where the restoring force is proportional to displacement, commonly modeled by mass-spring systems or pendulums. In this scenario, a 0.60 kg cart on a spring with k=240 N/m oscillates with amplitude A=0.070 m, and we need the maximum acceleration amax=ω²A. Choice A is correct because ω=√(k/m)=√(240/0.60)=20 rad/s, giving amax=(20)²×0.070=400×0.070=28 m/s². Choice B is incorrect due to a common error where students might calculate amax=ωA instead of ω²A, getting 20×0.070=1.4 m/s² (though this doesn't match choice B exactly). To help students: Emphasize that acceleration in SHM is proportional to displacement with factor -ω², making maximum acceleration occur at maximum displacement. Practice problems should include finding acceleration at various positions to reinforce the a=-ω²x relationship.
A mass attached to a spring oscillates with simple harmonic motion. At time $t = 0$ , the mass is at position $x = +A/2$ and moving in the positive direction with speed $v = \frac{\sqrt{3}}{2}\omega A$ , where $A$ is the amplitude and $\omega$ is the angular frequency. What is the phase constant $\phi$ in the equation $x(t) = A\cos(\omega t + \phi)$ ?
Answer: $-\frac{\pi}{3}$ Given $x(t) = A\cos(\omega t + \phi)$ , we have $v(t) = -A\omega\sin(\omega t + \phi)$ . At $t = 0$ : $x(0) = A\cos(\phi) = A/2$ , so $\cos(\phi) = 1/2$ . Also, $v(0) = -A\omega\sin(\phi) = \frac{\sqrt{3}}{2}\omega A$ , so $\sin(\phi) = -\frac{\sqrt{3}}{2}$ . Since $\cos(\phi) = 1/2 > 0$ and $\sin(\phi) = -\frac{\sqrt{3}}{2} < 0$ , $\phi$ is in the fourth quadrant, giving $\phi = -\frac{\pi}{3}$ . Choice B gives the wrong sign. Choices C and D correspond to $\sin(\phi) = \pm\frac{1}{2}$ instead of $\pm\frac{\sqrt{3}}{2}$ .
A simple pendulum has length $L=0.80\ \text{m}$ and bob mass $m=0.15\ \text{kg}$ , oscillating with small angle so SHM applies. Take $g=9.8\ \text{m/s}^2$ and equilibrium at $\theta=0$ . The angular frequency is $\omega=\sqrt{g/L}$ and the motion can be written $\theta(t)=\theta_{\max}\cos(\omega t+\phi)$ with $\theta$ in radians. The period is $T=2\pi/\omega$ and the frequency is $f=1/T$ . The phase $\phi$ is set by initial conditions, but does not affect $T$ . Air resistance is negligible, so amplitude stays constant. What is the period $T$ of the pendulum?
Answer: $T=1.8\ \text{s}$ This question tests the ability to represent and analyze simple harmonic motion (SHM) in the context of AP Physics C: Mechanics. SHM is characterized by periodic motion where the restoring force is proportional to displacement, commonly modeled by mass-spring systems or pendulums. In this scenario, a simple pendulum with length L=0.80 m undergoes small-angle oscillations where the angular frequency is ω=√(g/L)=√(9.8/0.80)=3.5 rad/s. Choice B is correct because the period T=2π/ω=2π/3.5≈1.8 s, noting that the mass of the bob does not affect the period for small angles. Choice A is incorrect due to a common error where students might calculate T=π/ω instead of 2π/ω, getting half the correct period. To help students: Emphasize that pendulum period depends only on length and gravity for small angles, not on mass. Practice problems should include comparing pendulums of different masses but same length to reinforce this counterintuitive concept.
A 0.30 kg mass on a spring with $k=120\ \text{N/m}$ oscillates without friction about equilibrium $x=0$ . At $t=0$ , the mass passes through equilibrium moving in the positive direction with speed $0.60\ \text{m/s}$ . Using $x(t)=A\cos(\omega t+\phi)$ and $v(t)=-A\omega\sin(\omega t+\phi)$ with $\omega=\sqrt{k/m}$ , the condition $x(0)=0$ implies $\cos\phi=0$ . The sign of $v(0)$ sets the phase quadrant. All values are SI units and amplitude is constant. What is the phase constant $\phi$ ?
Answer: $\phi=-\pi/2\ \text{rad}$ This question tests the ability to represent and analyze simple harmonic motion (SHM) in the context of AP Physics C: Mechanics. SHM is characterized by periodic motion where the restoring force is proportional to displacement, commonly modeled by mass-spring systems or pendulums. In this scenario, a mass passes through equilibrium (x=0) at t=0 with positive velocity, requiring us to find the phase constant φ from initial conditions x(0)=0 and v(0)>0. Choice C is correct because x(0)=A cos(φ)=0 means cos(φ)=0, so φ=±π/2, and v(0)=-Aω sin(φ)>0 requires sin(φ)<0, which is satisfied by φ=-π/2. Choice A is incorrect because φ=π/2 would give sin(φ)=1, making v(0)=-Aω<0, contradicting the positive initial velocity. To help students: Emphasize the importance of checking both position and velocity conditions when determining phase. Practice problems should include various initial conditions to build intuition about how phase affects the starting point of oscillation.
A small-angle pendulum has length $L=1.2\ \text{m}$ and bob mass $m=0.20\ \text{kg}$ , swinging about $\theta=0$ with constant amplitude $\theta_{\max}$ . Take $g=9.8\ \text{m/s}^2$ and model the motion as $\theta(t)=\theta_{\max}\cos(\omega t+\phi)$ with $\omega=\sqrt{g/L}$ . The period is $T=2\pi/\omega$ and the frequency is $f=1/T$ . The phase $\phi$ depends on initial conditions but does not change $f$ . Assume no air resistance and SI units. Determine the frequency $f$ .
Answer: $f=0.45\ \text{Hz}$ This question tests the ability to represent and analyze simple harmonic motion (SHM) in the context of AP Physics C: Mechanics. SHM is characterized by periodic motion where the restoring force is proportional to displacement, commonly modeled by mass-spring systems or pendulums. In this scenario, a pendulum with length L=1.2 m undergoes small-angle oscillations with angular frequency ω=√(g/L)=√(9.8/1.2)=2.86 rad/s. Choice A is correct because the frequency f=ω/(2π)=2.86/(2π)≈0.455 Hz, which rounds to 0.45 Hz. Choice D is incorrect due to a common error where students might calculate f=1/T but use an incorrect period, possibly forgetting the 2π factor or misapplying the pendulum formula. To help students: Emphasize that pendulum frequency depends on √(g/L) and is independent of mass and amplitude for small angles. Practice problems should include comparing pendulums of different lengths to build intuition about the inverse square root relationship.
A horizontal spring-mass system has a block of mass $m$ attached to a spring with constant $k$ . The block is displaced from equilibrium and released. If the coefficient of kinetic friction between the block and surface is $\mu_k$ , which expression best represents the motion during the first quarter period after release?
Answer: The motion is not simple harmonic due to the non-conservative friction force Kinetic friction provides a constant force $f_k = \mu_k mg$ opposing motion, which changes direction each time the velocity changes sign. This makes the friction force non-conservative and velocity-dependent in direction. The equation of motion becomes $m\ddot{x} = -kx \mp \mu_k mg$ (sign depends on velocity direction), which is not the standard SHM equation $m\ddot{x} = -kx$ . Choice A incorrectly treats friction as shifting equilibrium. Choice B suggests exponential damping, but kinetic friction doesn't provide viscous damping. Choice C incorrectly modifies the frequency. The motion is actually piecewise harmonic with decreasing amplitude.
A 0.40 kg block attached to a spring with $k=160\ \text{N/m}$ oscillates on a frictionless surface about $x=0$ . At $t=0$ it is at $x=+0.050\ \text{m}$ moving in the negative direction with speed $0.40\ \text{m/s}$ . For SHM, $x(t)=A\cos(\omega t+\phi)$ with $\omega=\sqrt{k/m}$ and $v(t)=-A\omega\sin(\omega t+\phi)$ . The amplitude satisfies $A^2=x_0^2+(v_0/\omega)^2$ using SI units. The phase $\phi$ is determined by $x_0$ and $v_0$ , but $A$ is independent of sign conventions. Mechanical energy is conserved. What is the amplitude $A$ ?
Answer: $A=0.064\ \text{m}$ This question tests the ability to represent and analyze simple harmonic motion (SHM) in the context of AP Physics C: Mechanics. SHM is characterized by periodic motion where the restoring force is proportional to displacement, commonly modeled by mass-spring systems or pendulums. In this scenario, a 0.40 kg block on a spring with k=160 N/m has initial position x₀=0.050 m and initial velocity v₀=-0.40 m/s, requiring us to find the amplitude using A²=x₀²+(v₀/ω)². Choice B is correct because ω=√(k/m)=√(160/0.40)=20 rad/s, giving A²=(0.050)²+(0.40/20)²=0.0025+0.0004=0.0029, so A≈0.064 m. Choice C is incorrect due to a common error where students might forget to square the terms in the amplitude formula, calculating A=x₀+v₀/ω instead. To help students: Emphasize that amplitude represents the maximum displacement and can be found from initial conditions using energy conservation or the Pythagorean-like formula. Practice problems should include various combinations of initial position and velocity to reinforce this concept.
A 0.50 kg block on a frictionless track is attached to a spring with $k=200\ \text{N/m}$ and equilibrium at $x=0$ . It is pulled to $x=+0.080\ \text{m}$ and released from rest at $t=0$ , so $A=0.080\ \text{m}$ and $\phi=0$ . The angular frequency is $\omega=\sqrt{k/m}$ and the motion is $x(t)=A\cos(\omega t+\phi)$ with $x$ in meters and $t$ in seconds. The velocity is $v(t)=-A\omega\sin(\omega t+\phi)$ and the acceleration is $a(t)=-\omega^2 x(t)$ . The restoring force is $F_s=-kx$ . Assume SHM holds for all times. What is the period $T$ of the oscillation?
Answer: $T=0.31\ \text{s}$ This question tests the ability to represent and analyze simple harmonic motion (SHM) in the context of AP Physics C: Mechanics. SHM is characterized by periodic motion where the restoring force is proportional to displacement, commonly modeled by mass-spring systems or pendulums. In this scenario, a 0.50 kg block attached to a spring with k=200 N/m undergoes SHM with angular frequency ω=√(k/m)=√(200/0.50)=20 rad/s. Choice B is correct because the period T=2π/ω=2π/20≈0.314 s, which rounds to 0.31 s. Choice D is incorrect due to a common error where students might calculate T=2π√(m/k) but forget the 2π factor, getting approximately 0.16 s instead. To help students: Emphasize the importance of remembering the complete period formula T=2π/ω and the relationship between angular frequency and spring-mass parameters. Practice problems should include variations in mass and spring constant to reinforce the inverse relationship between period and angular frequency.

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