Energy of Simple Harmonic Oscillators - AP Physics C Mechanics

1. What is the ratio of the total mechanical energy of System 2 to that of System 1, $E_2/E_1$?
   Answer: 2 The total energy of a mass-spring oscillator is given by $E = \frac{1}{2}kA^2$. For System 1, $E_1 = \frac{1}{2}kA^2$. For System 2, $E_2 = \frac{1}{2}k_2A_2^2 = \frac{1}{2}(k/2)(2A)^2 = \frac{1}{2}(k/2)(4A^2) = 2(\frac{1}{2}kA^2) = 2E_1$. The ratio $E_2/E_1$ is 2.
2. A physical pendulum consists of a rigid object of mass $M$ that oscillates about a pivot point a distance $d$ from its center of mass. It is released from rest at a small maximum angular displacement $\theta_{max}$. Which of the following changes will increase the total mechanical energy of the pendulum-Earth system?
   Answer: Increasing the mass $M$ of the pendulum while keeping $d$ and $\theta_{max}$ constant. The total mechanical energy is determined by the maximum gravitational potential energy. The maximum height of the center of mass above its equilibrium position is $h = d(1-\cos\theta_{max})$. The total energy is $E = Mgh = Mgd(1-\cos\theta_{max})$. To increase $E$, one must increase $M$, $g$, $d$, or $\theta_{max}$. Increasing the mass $M$ while other factors are constant will increase the total energy.
3. What is the speed of the object when its displacement from equilibrium is 0.30 m?
   Answer: 0.80 m/s The total energy is $E = \frac{1}{2}kA^2 = \frac{1}{2}(8.0)(0.50)^2 = 1.0$ J. At any position $x$, the total energy is the sum of kinetic and potential energy: $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$. At $x=0.30$ m, $1.0 = \frac{1}{2}(2.0)v^2 + \frac{1}{2}(8.0)(0.30)^2$. This simplifies to $1.0 = v^2 + 0.36$, so $v^2 = 0.64$, and $v = 0.80$ m/s.
4. An object of mass $m$ is in simple harmonic motion with total energy $E$. If the total energy is increased to $4E$ while the mass remains constant, by what factor does the maximum speed of the object change?
   Answer: It increases by a factor of 2. The total energy $E$ is equal to the maximum kinetic energy, $E = \frac{1}{2}mv_{max}^2$. Therefore, $v_{max} = \sqrt{2E/m}$. The maximum speed is proportional to the square root of the total energy. If the energy is increased by a factor of 4, the maximum speed will increase by a factor of $\sqrt{4} = 2$.
5. A particle executes SHM with amplitude $A$ and total energy $E$. If the amplitude is reduced to $\frac{A}{2}$ while the mass and spring constant remain unchanged, what additional work must be done by an external agent to restore the total energy to $E$?
   Answer: $\frac{3E}{4}$ When you encounter SHM energy problems involving amplitude changes, focus on the relationship between total energy and amplitude. In simple harmonic motion, the total energy is $E = \frac{1}{2}kA^2$, where $k$ is the spring constant and $A$ is the amplitude. Initially, the particle has energy $E$ with amplitude $A$. When the amplitude is reduced to $\frac{A}{2}$, the new energy becomes $E_{new} = \frac{1}{2}k(\frac{A}{2})^2 = \frac{1}{2}k \cdot \frac{A^2}{4} = \frac{1}{4} \cdot \frac{1}{2}kA^2 = \frac{E}{4}$. To restore the total energy back to $E$, an external agent must add energy equal to the difference: $E - \frac{E}{4} = \frac{3E}{4}$. This makes choice A correct. Choice B ($\frac{E}{2}$) incorrectly assumes the energy scales linearly with amplitude rather than quadratically. Choice C ($\frac{E}{4}$) gives the remaining energy after the amplitude reduction, not the work needed to restore the original energy. Choice D ($E$) would be the work needed if all energy were lost, ignoring that $\frac{E}{4}$ remains in the system. Remember that in SHM, energy depends on the square of the amplitude ($E \propto A^2$). When solving energy problems involving amplitude changes, always calculate the energy difference between final and initial states. This quadratic relationship is crucial for understanding how energy storage changes in oscillatory systems.
6. An object of mass $m$ is attached to an ideal horizontal spring with spring constant $k$. The object is displaced from its equilibrium position by a distance $A$ and released from rest. Assuming no friction, what is the total mechanical energy of the object-spring system?
   Answer: $E = \frac{1}{2}kA^2$ The total mechanical energy of a simple harmonic oscillator is constant. At the point of maximum displacement (the amplitude $A$), the object is momentarily at rest, so its kinetic energy is zero. At this point, all the energy is stored as potential energy in the spring, which is given by $U_s = \frac{1}{2}kA^2$. Therefore, the total mechanical energy is $E = \frac{1}{2}kA^2$.
7. A frictionless spring-mass system has $k=200\,\text{N/m}$ and amplitude $A=0.10\,\text{m}$. At what displacement magnitude $|x|$ is $K=U$?
   Answer: $0.071\,\text{m}$ This question tests AP Physics C: Mechanics understanding of energy transformations in simple harmonic oscillators, specifically finding where kinetic and potential energies are equal. When K = U, each equals half the total energy since E = K + U. Setting U = E/2: ½kx² = ½(½kA²), which simplifies to x² = A²/2, giving |x| = A/√2 ≈ 0.707A. With A = 0.10 m, |x| = 0.10/√2 ≈ 0.0707 m ≈ 0.071 m. Choice B is correct because it matches this calculation. Choice C (0.050 m = A/2) is a common error from assuming linear rather than quadratic energy relationships. To help students: emphasize that energy varies as x², practice solving for positions where K and U have specific ratios, and use energy bar charts to visualize energy distribution at different positions.
8. Two mass-spring systems have the same total energy $E$ but different masses: $m_1 = m$ and $m_2 = 4m$. Both systems have identical spring constants. The ratio of the maximum kinetic energy of system 1 to the maximum kinetic energy of system 2 is:
   Answer: $1$ When you encounter mass-spring systems with identical total energies, remember that energy conservation principles govern the relationship between kinetic and potential energy throughout the oscillation. For any mass-spring system, the total energy equals the maximum kinetic energy, which occurs when the spring passes through equilibrium (where potential energy is zero). Since both systems have the same total energy $E$, they must also have the same maximum kinetic energy. Here's why: At maximum displacement, all energy is potential ($E = \frac{1}{2}kA^2$). At equilibrium, all energy is kinetic ($E = \frac{1}{2}mv_{max}^2$). Since total energy is conserved and identical for both systems, $KE_{max,1} = KE_{max,2} = E$. Therefore, the ratio is $\frac{KE_{max,1}}{KE_{max,2}} = \frac{E}{E} = 1$. Choice A ($\frac{1}{4}$) incorrectly assumes maximum kinetic energy is inversely proportional to mass. Choice B ($\frac{1}{2}$) might come from incorrectly relating the mass ratio. Choice C ($4$) reverses the expected relationship, perhaps thinking heavier masses have less kinetic energy. The key insight is that while the masses affect the maximum velocities ($v_{max} = \sqrt{\frac{2E}{m}}$), the constraint of equal total energies ensures equal maximum kinetic energies regardless of mass differences. Study tip: For oscillating systems, always identify what's being held constant. When total energy is fixed, maximum kinetic energy equals that total energy, independent of mass or spring constant.
9. A mass oscillates on a spring with total energy $E$. At a certain instant, the potential energy is $\frac{E}{3}$ and the mass is moving away from equilibrium. After what fraction of a complete period will the potential energy again equal $\frac{E}{3}$?
   Answer: $\frac{1}{3}$ This problem tests your understanding of simple harmonic motion and energy conservation. When analyzing oscillatory motion, remember that total energy remains constant while kinetic and potential energy interchange periodically. Since total energy $E = KE + PE$ and is conserved, when $PE = \frac{E}{3}$, the kinetic energy must be $KE = \frac{2E}{3}$. The potential energy in simple harmonic motion varies as $PE = \frac{1}{2}kx^2$, where the maximum potential energy equals the total energy $E$ at the turning points. When $PE = \frac{E}{3}$, the mass is at a specific displacement from equilibrium. Since the mass is moving away from equilibrium, it will continue until it reaches the turning point where $PE = E$. Then it returns, passing through the same position (where $PE = \frac{E}{3}$ again) as it moves back toward equilibrium. Due to the symmetry of simple harmonic motion, if you start at any position and move away from equilibrium, you'll return to that same position after traveling to the turning point and back. This journey takes exactly one-third of a complete period. Answer choice A ($\frac{1}{6}$) represents the time to reach maximum displacement only. Choice B ($\frac{1}{4}$) is the time between equilibrium and maximum displacement, which doesn't apply here. Choice C ($\frac{2}{3}$) would be twice the correct time interval. Study tip: For SHM energy problems, sketch the position-time graph and mark where equal potential energies occur. The symmetry of oscillatory motion often provides the key insight for timing questions.
10. A simple pendulum consists of a bob of mass $m$ attached to a string of length $L$. The pendulum is pulled back to a maximum angle $\theta_{max}$ with the vertical and released from rest. What is the total mechanical energy of the pendulum-Earth system with respect to the lowest point of the swing?
    Answer: $E = mgL(1 - \cos\theta_{max})$ The total mechanical energy is conserved. At the maximum angular displacement, the bob is momentarily at rest, so its kinetic energy is zero. The total energy is equal to the gravitational potential energy at that point. The height of the bob above the lowest point is $h = L - L\cos\theta_{max} = L(1 - \cos\theta_{max})$. Therefore, the total energy is $E = mgh = mgL(1 - \cos\theta_{max})$.
11. A mass attached to a vertical spring oscillates with amplitude $A$. Taking the equilibrium position as the reference for gravitational potential energy, at what displacement from equilibrium is the total mechanical energy (including gravitational potential energy) equal to twice the elastic potential energy?
    Answer: $\pm \frac{A}{\sqrt{2}}$ Total mechanical energy $E = \frac{1}{2}kA^2$. Elastic potential energy at displacement $x$ is $U_e = \frac{1}{2}kx^2$. When $E = 2U_e$: $\frac{1}{2}kA^2 = 2 \cdot \frac{1}{2}kx^2$, so $A^2 = 2x^2$, giving $x = \pm \frac{A}{\sqrt{2}}$. Note that gravitational PE is zero at equilibrium by the problem setup. Choice B uses $A^2 = 4x^2$. Choice C comes from incorrectly using energy ratios. Choice D inverts the square root.
12. An object of mass $m$ is attached to a spring and undergoes simple harmonic motion with total energy $E$. Which of the following expressions correctly gives the maximum speed, $v_{max}$, of the object?
    Answer: $v_{max} = \sqrt{2E/m}$ The total mechanical energy $E$ is constant. At the equilibrium position, the potential energy is zero and the kinetic energy is maximum. Therefore, the total energy is equal to the maximum kinetic energy: $E = K_{max} = \frac{1}{2}mv_{max}^2$. Solving for $v_{max}$ gives $v_{max} = \sqrt{2E/m}$.
13. For an undamped simple harmonic oscillator with total mechanical energy $E$, what is the time-average value of the potential energy over one complete period of oscillation?
    Answer: $E/2$ Over a full cycle of simple harmonic motion, energy is continuously exchanged between kinetic and potential forms. The time-average value of the kinetic energy is equal to the time-average value of the potential energy. Since their sum must be the constant total energy $E$, each must have an average value of $E/2$. So, $<U> = <K> = E/2$.
14. A block attached to a spring oscillates horizontally with total mechanical energy $E$. What is the kinetic energy of the block when its displacement is half the amplitude, $A/2$?
    Answer: $3E/4$ The total energy is $E = \frac{1}{2}kA^2$. The potential energy at displacement $x$ is $U = \frac{1}{2}kx^2$. At $x = A/2$, the potential energy is $U = \frac{1}{2}k(A/2)^2 = \frac{1}{4}(\frac{1}{2}kA^2) = E/4$. By conservation of energy, the kinetic energy is $K = E - U = E - E/4 = 3E/4$.
15. A mass-spring system oscillates with simple harmonic motion. When the displacement is $\frac{1}{3}$ of the amplitude, the ratio of kinetic energy to potential energy is:
    Answer: $\frac{8}{1}$ For SHM, $U = \frac{1}{2}kx^2$ and $K = \frac{1}{2}k(A^2 - x^2)$. When $x = \frac{A}{3}$: $U = \frac{1}{2}k(\frac{A}{3})^2 = \frac{kA^2}{18}$ and $K = \frac{1}{2}k(A^2 - \frac{A^2}{9}) = \frac{1}{2}k \cdot \frac{8A^2}{9} = \frac{4kA^2}{9}$. Therefore $\frac{K}{U} = \frac{\frac{4kA^2}{9}}{\frac{kA^2}{18}} = \frac{4kA^2}{9} \cdot \frac{18}{kA^2} = 8$. Choice A uses $\frac{U}{K}$ instead. Choice B confuses the calculation. Choice D incorrectly squares the amplitude ratio.
16. How does the new total mechanical energy $E_{new}$ compare to the original total mechanical energy $E_{orig}$?
    Answer: $E_{new} = 9 E_{orig}$ The total mechanical energy of a mass-spring system in simple harmonic motion is given by $E = \frac{1}{2}kA^2$, where $k$ is the spring constant and $A$ is the amplitude. Since the energy is proportional to the square of the amplitude, tripling the amplitude ($A_{new} = 3A_{orig}$) results in the new energy being $3^2 = 9$ times the original energy.
17. The potential energy of a particle undergoing one-dimensional simple harmonic motion is given by the function $U(x) = (2.0 \text{ J/m}^2)x^2$. If the total mechanical energy of the particle is 8.0 J, what is the amplitude of the oscillation?
    Answer: 2.0 m The total mechanical energy $E$ is equal to the maximum potential energy, which occurs at the amplitude, $x=A$. So, $E = U(A)$. We have $8.0 \text{ J} = (2.0 \text{ J/m}^2)A^2$. Solving for $A^2$ gives $A^2 = 8.0/2.0 = 4.0 \text{ m}^2$. Taking the square root gives the amplitude $A = 2.0$ m.
18. A particle of mass $m$ is attached to a spring with constant $k$ and is undergoing simple harmonic motion. At a displacement $x_1$ from equilibrium, the particle has a speed $v_1$. Which of the following expressions represents the total mechanical energy of the particle-spring system?
    Answer: $\frac{1}{2}mv_1^2 + \frac{1}{2}kx_1^2$ The total mechanical energy $E$ of a conservative system is the sum of its kinetic energy $K$ and potential energy $U$. At any point in the motion, $E = K + U$. For a mass-spring system, the kinetic energy is $\frac{1}{2}mv^2$ and the elastic potential energy is $\frac{1}{2}kx^2$. Thus, at the given instant, the total energy is $\frac{1}{2}mv_1^2 + \frac{1}{2}kx_1^2$.
19. A mass-spring system is oscillating with amplitude $A$ and total energy $E$. The motion is subject to a small damping force. Which of the following correctly describes the energy of the system after a long time?
    Answer: The total mechanical energy gradually approaches zero. A damping force is a non-conservative force that does negative work on the system. This work removes mechanical energy from the system, typically converting it into thermal energy. As a result, the amplitude of oscillation decreases, and the total mechanical energy of the system gradually decreases, approaching zero as the system comes to rest at its equilibrium position.
20. Two identical masses are attached to springs with different spring constants $k_1$ and $k_2$ where $k_2 = 4k_1$. Both systems have the same total mechanical energy. The ratio of the amplitude of oscillation for system 1 to system 2 is:
    Answer: $2$ Total energy is $E = \frac{1}{2}kA^2$. Since energies are equal: $\frac{1}{2}k_1A_1^2 = \frac{1}{2}k_2A_2^2$, so $k_1A_1^2 = k_2A_2^2$. Substituting $k_2 = 4k_1$: $k_1A_1^2 = 4k_1A_2^2$, which gives $A_1^2 = 4A_2^2$, so $\frac{A_1}{A_2} = 2$. Choice A assumes direct proportionality. Choice B takes the square root incorrectly. Choice D assumes inverse relationship without the square root.

Start this quiz