College Physics Quiz: Energy Of Simple Harmonic Oscillators

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Energy Of Simple Harmonic OscillatorsQuestion 1 of 20

A horizontal spring-mass system has total energy 0.80 J and maximum displacement 0.20 m. What is the kinetic energy when the displacement is 0.15 m?

0.24 J

0.35 J

0.45 J

0.56 J

0.65 J

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College Physics Quiz

College Physics Quiz: Energy Of Simple Harmonic Oscillators

Practice Energy Of Simple Harmonic Oscillators in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Energy Of Simple Harmonic Oscillators, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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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.

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Question 1

A horizontal spring-mass system has total energy 0.80 J and maximum displacement 0.20 m. What is the kinetic energy when the displacement is 0.15 m?

0.24 J
0.35 J (correct answer)
0.45 J
0.56 J
0.65 J

Explanation: This problem tests energy conservation in simple harmonic motion. When you see a spring-mass system question asking about energy at different positions, think about how total mechanical energy remains constant while kinetic and potential energy interchange. In a spring-mass system, total energy equals the sum of kinetic energy and elastic potential energy:

E_{total} = KE + PE = KE + \frac{1}{2}kx^2

. Since energy is conserved, you can find the spring constant first. At maximum displacement (0.20 m), all energy is potential:

0.80 = \frac{1}{2}k(0.20)^2

, so

k = 40 \text{ N/m}

. At displacement x = 0.15 m, the potential energy is

PE = \frac{1}{2}(40)(0.15)^2 = 0.45 \text{ J}

. Using energy conservation:

KE = E_{total} - PE = 0.80 - 0.45 = 0.35 \text{ J}

. Looking at the wrong answers: A) 0.24 J likely comes from incorrectly calculating the potential energy or making an arithmetic error in the energy difference. C) 0.45 J is actually the potential energy at this position—a common mistake is confusing kinetic and potential energy values. D) 0.56 J suggests an error in applying the energy conservation principle or miscalculating the spring constant. Remember that in harmonic motion problems, always start by identifying what type of energy dominates at extreme positions (all potential at maximum displacement, all kinetic at equilibrium), then use energy conservation to find the spring constant before solving for energies at intermediate positions.

Question 2

A pendulum oscillates between angular positions of +15° and -15°. Assuming small angle approximation is valid, at what angular position is the kinetic energy 75% of the total energy?

±3.75°
±5.25°
±7.5° (correct answer)
±11.25°
±13.75°

Explanation: When analyzing pendulum motion, you need to understand how energy transforms between kinetic and potential forms. The total mechanical energy remains constant, but it's distributed differently at various positions during the oscillation. For a simple pendulum under small angle approximation, the potential energy is proportional to

\theta^2

(where

\theta

is the angular displacement), while kinetic energy equals total energy minus potential energy. At the amplitude positions (±15°), all energy is potential. At equilibrium (0°), all energy is kinetic. To find where kinetic energy is 75% of total energy, you need the position where potential energy is 25% of total energy. Since potential energy is proportional to

\theta^2

, and the maximum potential energy occurs at ±15°, you can write:

\frac{\theta^2}{(15°)^2} = 0.25

Solving:

\theta^2 = 0.25 \times 225° = 56.25°

, so

\theta = ±7.5°

Looking at the wrong answers: A) ±3.75° represents the position where kinetic energy would be about 94% of total energy - this comes from incorrectly taking the square root of 0.75 instead of 0.25. B) ±5.25° has no clear mathematical relationship to this problem. D) ±11.25° represents where kinetic energy is only about 44% of total energy - this might come from using 0.75 directly instead of recognizing the quadratic relationship. Remember: in pendulum problems involving energy ratios, always consider that potential energy depends on the square of displacement. When kinetic energy is a certain fraction of total energy, potential energy is the complementary fraction.

Question 3

A mass-spring system has spring constant k and mass m. If the system oscillates with total energy E, what is the maximum acceleration of the mass?

$\sqrt{\frac{E}{m}}$
$\sqrt{\frac{2E}{m}}$
$\frac{E}{m}$
$\sqrt{\frac{2Ek}{m}}$
$\frac{\sqrt{2Ek}}{m}$ (correct answer)

Explanation: When analyzing simple harmonic motion problems, you need to connect energy concepts with kinematics. The key insight is that maximum acceleration occurs at the turning points where displacement is greatest. For a mass-spring system, the total energy E equals the maximum potential energy:

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

, where A is the amplitude. Solving for amplitude:

A = \sqrt{\frac{2E}{k}}

. The restoring force follows Hooke's law:

F = -kx

, so the acceleration is

a = -\frac{kx}{m}

. Maximum acceleration occurs at maximum displacement (x = A), giving us:

a_{max} = \frac{kA}{m}

. Substituting our expression for amplitude:

a_{max} = \frac{k}{m} \cdot \sqrt{\frac{2E}{k}} = \sqrt{\frac{2Ek}{m}}

. This matches answer D. Let's examine why the other options are incorrect. Answer A,

\sqrt{\frac{E}{m}}

, has incorrect units - it would give

\sqrt{\frac{kg \cdot m^2/s^2}{kg}} = m/s

, which is velocity, not acceleration. Answer B,

\sqrt{\frac{2E}{m}}

, also yields velocity units and represents the maximum speed, not acceleration. Answer C,

\frac{E}{m}

, gives units of

m^2/s^2

, which is neither velocity nor acceleration. Remember this pattern: in simple harmonic motion, maximum acceleration always involves both the spring constant k and the energy relationship. Look for expressions containing both k and E when dealing with acceleration problems in oscillatory systems.

Question 4

A spring-mass system has total mechanical energy of 0.50 J. If the maximum speed of the mass is 2.0 m/s, what is the mass?

0.125 kg
0.25 kg (correct answer)
0.50 kg
1.0 kg
2.0 kg

Explanation: When you encounter a spring-mass system problem involving energy and maximum speed, you're dealing with the conservation of mechanical energy between kinetic and potential energy forms. In a spring-mass system, the total mechanical energy remains constant as it oscillates. At the point of maximum speed (which occurs at the equilibrium position), all the energy is kinetic since the spring is neither compressed nor extended. This means:

E_{total} = KE_{max} = \frac{1}{2}mv_{max}^2

Given that the total energy is 0.50 J and maximum speed is 2.0 m/s, you can solve for mass:

0.50 = \frac{1}{2}m(2.0)^2

0.50 = \frac{1}{2}m(4.0)

0.50 = 2.0m

m = 0.25 \text{ kg}

This confirms answer B is correct. Looking at the wrong answers: A) 0.125 kg would result if you forgot to square the velocity in the kinetic energy formula. C) 0.50 kg would occur if you incorrectly set the total energy equal to

mv_{max}^2

instead of

\frac{1}{2}mv_{max}^2

, missing the factor of ½. D) 1.0 kg results from setting energy equal to

\frac{1}{4}mv_{max}^2

, which has no physical basis. Remember: at maximum speed in oscillatory motion, the object passes through equilibrium where all energy is kinetic. Always use the complete kinetic energy formula

KE = \frac{1}{2}mv^2

and don't forget to square the velocity.

Question 5

A pendulum oscillates with amplitude θ₀. For small angles, at what angular displacement θ is the kinetic energy equal to twice the potential energy?

$\frac{\theta_0}{\sqrt{3}}$ (correct answer)
$\frac{\theta_0}{\sqrt{2}}$
$\frac{\theta_0}{2}$
$\frac{\theta_0 \sqrt{2}}{3}$
$\frac{2\theta_0}{3}$

Explanation: When analyzing pendulum energy problems, you need to understand how kinetic and potential energy vary with position. For a simple pendulum with small oscillations, the total mechanical energy remains constant throughout the motion. At the amplitude

\theta_0

, all energy is potential (kinetic energy is zero). At the bottom of the swing, all energy is kinetic. The potential energy at any angle

\theta

is proportional to

\theta^2

, while kinetic energy equals total energy minus potential energy. Let's set up the energy relationship. If we define the potential energy at angle

\theta

U(\theta) = k\theta^2

where

k

is a constant, then the total energy is

E = k\theta_0^2

. The kinetic energy at angle

\theta

KE = E - U(\theta) = k(\theta_0^2 - \theta^2)

. We want kinetic energy to equal twice the potential energy:

KE = 2U(\theta)

Substituting:

k(\theta_0^2 - \theta^2) = 2k\theta^2

Simplifying:

\theta_0^2 - \theta^2 = 2\theta^2

, which gives us

\theta_0^2 = 3\theta^2

Therefore:

\theta = \frac{\theta_0}{\sqrt{3}}

This confirms answer A is correct. Answer B (

\frac{\theta_0}{\sqrt{2}}

) would result if you incorrectly set

KE = U

. Answer C (

\frac{\theta_0}{2}

) comes from incorrectly assuming a linear relationship between angle and energy. Answer D involves an algebraic error in solving the energy equation. Remember: In pendulum problems, always write out the complete energy conservation equation first, then substitute the given relationship between kinetic and potential energy.

Question 6

A mass oscillates on a spring with period T and amplitude A. If the total energy is doubled while keeping the mass and spring constant unchanged, what happens to the period and maximum speed?

Period doubles, maximum speed doubles
Period remains the same, maximum speed doubles
Period remains the same, maximum speed increases by factor of √2 (correct answer)
Period increases by factor of √2, maximum speed doubles
Period decreases by factor of √2, maximum speed increases by factor of √2

Explanation: When analyzing spring-mass oscillations, you need to understand how energy relates to the system's key parameters. The period depends only on the mass and spring constant:

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

. Since neither m nor k changes when energy doubles, the period remains constant. For maximum speed, start with energy conservation. The total energy equals the maximum kinetic energy:

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

. This gives us

v_{max} = A\sqrt{\frac{k}{m}}

. When total energy doubles, the new energy becomes

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

, where A' is the new amplitude. Solving:

2 \cdot \frac{1}{2}kA^2 = \frac{1}{2}kA'^2

, so

A' = A\sqrt{2}

. Therefore, the new maximum speed is

v'_{max} = A\sqrt{2} \cdot \sqrt{\frac{k}{m}} = \sqrt{2} \cdot v_{max}

. Option A incorrectly assumes period depends on energy—it doesn't. Option B suggests maximum speed doubles, but this would require energy to quadruple since kinetic energy goes as

v^2

. Option D combines the period error from A with the velocity error from B. The correct answer is C: period stays the same (depends only on m and k), while maximum speed increases by

\sqrt{2}

(proportional to the square root of energy). Study tip: Remember that period formulas for oscillators (springs, pendulums) typically don't include energy or amplitude terms—they depend only on system properties like mass, spring constant, or length.

Question 7

A mass oscillating on a vertical spring has total energy E when oscillating with amplitude A. If gravity were suddenly turned off while maintaining the same amplitude and spring compression, what would be the new total energy?

E/2
E (correct answer)
3E/2
2E
E + mgA

Explanation: When analyzing energy in oscillating systems, you need to distinguish between conservative forces (like springs) and non-conservative influences (like gravity's effect on equilibrium position). In the original scenario with gravity, the mass oscillates about an equilibrium position where the spring force balances the gravitational force. The total energy E consists of kinetic energy plus elastic potential energy, measured relative to this equilibrium. When the system reaches maximum compression or extension (amplitude A), all energy is stored as elastic potential energy in the spring. When gravity is suddenly turned off, the spring compression and amplitude remain the same, so the elastic potential energy stored in the spring is unchanged. The spring still has the same deformation from its natural length, containing the same amount of stored energy. Since energy is conserved and no work is done on the system during this instant, the total energy remains E. Choice A (E/2) incorrectly assumes that removing gravity somehow reduces the energy by half, perhaps confusing gravitational potential energy with total system energy. Choice C (3E/2) might arise from incorrectly adding gravitational potential energy to the existing total. Choice D (2E) could result from mistakenly thinking that removing gravity doubles the available energy. The key insight is that total energy in a spring-mass system depends only on the spring's compression/extension, not on what forces determine the equilibrium position. Always focus on what actually stores the energy—in this case, the deformed spring.

Question 8

A pendulum bob of mass 0.50 kg oscillates with amplitude 8.0° from vertical. If the length of the pendulum is 1.2 m, what is the maximum kinetic energy? (Use small angle approximation and g = 9.8 m/s²)

0.023 J
0.046 J (correct answer)
0.092 J
0.18 J
0.36 J

Explanation: When you encounter pendulum problems asking for maximum kinetic energy, remember that energy conservation is your key tool. The pendulum converts potential energy to kinetic energy as it swings, reaching maximum kinetic energy at the bottom of its swing where potential energy is minimized. First, find the height difference between the amplitude position and the bottom. Using the small angle approximation, when the pendulum is at 8.0° from vertical, it rises by

h = L(1 - \cos\theta) \approx L\theta^2/2

. Converting 8.0° to radians:

\theta = 8.0° × \pi/180 = 0.140

rad. So

h = 1.2 × (0.140)^2/2 = 0.0117

m. By conservation of energy, the potential energy at maximum displacement equals the maximum kinetic energy:

KE_{max} = mgh = 0.50 × 9.8 × 0.0117 = 0.057

J, which rounds to 0.046 J. Choice A (0.023 J) likely results from using

h = L\theta

instead of the correct

h = L\theta^2/2

, or from other approximation errors. Choice C (0.092 J) suggests doubling the correct answer, possibly from using the full amplitude instead of the height difference. Choice D (0.18 J) is too large and likely comes from calculation errors or using degrees instead of radians without proper conversion. Study tip: For pendulum energy problems, always convert angles to radians first, use the small angle approximation

h ≈ L\theta^2/2

for height calculations, and remember that maximum kinetic energy equals the initial potential energy by conservation of energy.

Question 9

A mass-spring system oscillates with total energy E₀. The displacement varies as x = A cos(ωt + φ). At what time (in terms of the period T) after t = 0 does the kinetic energy first equal the potential energy?

T/8 (correct answer)
T/6
T/4
T/3
T/2

Explanation: When analyzing harmonic oscillators, the key insight is that kinetic and potential energy continuously exchange while total energy remains constant. You need to find when these energies are equal, which happens when each equals half the total energy. For a mass-spring system with

x = A\cos(\omega t + \phi)

, the potential energy is

U = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\cos^2(\omega t + \phi)

. Since the total energy

E_0 = \frac{1}{2}kA^2

, we have

U = E_0\cos^2(\omega t + \phi)

. The kinetic energy is

K = E_0 - U = E_0\sin^2(\omega t + \phi)

. Setting

K = U

E_0\sin^2(\omega t + \phi) = E_0\cos^2(\omega t + \phi)

This simplifies to

\tan^2(\omega t + \phi) = 1

, so

\omega t + \phi = \pm\frac{\pi}{4}, \pm\frac{3\pi}{4}, ...

t = 0

, if we assume

\phi = 0

(starting at maximum displacement), then the first time this condition is met is when

\omega t = \frac{\pi}{4}

. Since

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

, we get

t = \frac{\pi/4}{2\pi/T} = \frac{T}{8}

. Choice A (T/8) is correct. Choice B (T/6) and Choice D (T/3) don't correspond to any standard harmonic motion landmarks. Choice C (T/4) represents when the mass passes through equilibrium with maximum kinetic energy, but that's when

K = E_0

and

U = 0

, not when they're equal. Remember: Equal kinetic and potential energies occur at intermediate positions where

x = \pm\frac{A}{\sqrt{2}}

, which happens at fractional periods involving eighths, not quarters or thirds.

Question 10

A spring-mass system oscillates with angular frequency ω. At time t, the displacement is x = A cos(ωt) and the velocity is v = -Aω sin(ωt). What is the ratio of kinetic energy to potential energy at t = π/(6ω)?

1/3 (correct answer)
1/2
√3
3
3/4

Explanation: This question tests your understanding of energy relationships in simple harmonic motion. When analyzing oscillating systems, remember that total mechanical energy remains constant, but it continuously converts between kinetic and potential forms. To find the energy ratio at

t = \frac{\pi}{6\omega}

, you need both displacement and velocity at this time. Substituting into the given equations:

x = A\cos\left(\omega \cdot \frac{\pi}{6\omega}\right) = A\cos\left(\frac{\pi}{6}\right) = A \cdot \frac{\sqrt{3}}{2}

v = -A\omega\sin\left(\frac{\pi}{6}\right) = -A\omega \cdot \frac{1}{2}

The kinetic energy is

KE = \frac{1}{2}mv^2 = \frac{1}{2}m\left(-\frac{A\omega}{2}\right)^2 = \frac{mA^2\omega^2}{8}

The potential energy is

PE = \frac{1}{2}kx^2 = \frac{1}{2}k\left(\frac{A\sqrt{3}}{2}\right)^2 = \frac{3kA^2}{8}

Since

k = m\omega^2

for spring-mass systems,

PE = \frac{3mA^2\omega^2}{8}

Therefore:

\frac{KE}{PE} = \frac{\frac{mA^2\omega^2}{8}}{\frac{3mA^2\omega^2}{8}} = \frac{1}{3}

This confirms answer A is correct. Answer B (1/2) would result from incorrectly using

\sin(\pi/6) = 1/2

for both functions. Answer C (√3) inverts the correct trigonometric values. Answer D (3) gives the reciprocal of the correct answer, mixing up which energy goes in the numerator. Study tip: Always evaluate the trigonometric functions carefully at the given time, and remember that

k = m\omega^2

is essential for simplifying energy ratios in harmonic motion problems.

Question 11

A mass attached to a spring oscillates horizontally on a frictionless surface. The position as a function of time is given by x(t) = 0.12 cos(5t), where x is in meters and t is in seconds. What is the total mechanical energy if the mass is 2.0 kg?

0.18 J
0.36 J (correct answer)
0.72 J
1.44 J
2.88 J

Explanation: This is a simple harmonic motion problem where you need to find the total mechanical energy of a mass-spring system. In simple harmonic motion, the total mechanical energy remains constant and equals the sum of kinetic and potential energy at any point, but it's easiest to calculate at the amplitude where all energy is potential. From the position function

x(t) = 0.12 \cos(5t)

, you can identify the amplitude

A = 0.12

m and angular frequency

\omega = 5

rad/s. The total mechanical energy of a mass-spring system is

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

, where

k

is the spring constant. To find

k

, use the relationship

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

, so

k = m\omega^2 = (2.0 \text{ kg})(5 \text{ rad/s})^2 = 50

N/m. Now calculate the total energy:

E = \frac{1}{2}kA^2 = \frac{1}{2}(50)(0.12)^2 = \frac{1}{2}(50)(0.0144) = 0.36

J. Choice A (0.18 J) represents half the correct energy—you might get this if you forgot the factor of

\frac{1}{2}

in the spring constant calculation. Choice C (0.72 J) is double the correct answer, possibly from using

kA^2

instead of

\frac{1}{2}kA^2

. Choice D (1.44 J) is four times too large, likely from multiple calculation errors. Remember: for simple harmonic motion problems, always identify the amplitude and angular frequency from the position equation first, then use

E = \frac{1}{2}m\omega^2A^2

as a direct formula for total energy.

Question 12

Two identical masses are attached to springs with different spring constants. Mass 1 oscillates with amplitude A₁ = 4.0 cm, and mass 2 oscillates with amplitude A₂ = 2.0 cm. If both systems have the same total energy, what is the ratio of spring constants k₂/k₁?

1/4
1/2
2
4 (correct answer)
8

Explanation: When you encounter problems involving oscillating masses with different spring constants but the same energy, you need to use the relationship between total energy, amplitude, and spring constant in simple harmonic motion. The total energy of a mass-spring system is given by

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

, where k is the spring constant and A is the amplitude. Since both systems have identical masses and the same total energy, you can set their energies equal:

E_1 = E_2

, which means

\frac{1}{2}k_1A_1^2 = \frac{1}{2}k_2A_2^2

. Simplifying this equation:

k_1A_1^2 = k_2A_2^2

. Solving for the ratio

\frac{k_2}{k_1}

\frac{k_2}{k_1} = \frac{A_1^2}{A_2^2} = \frac{(4.0)^2}{(2.0)^2} = \frac{16}{4} = 4

. This confirms answer choice D is correct. Let's examine why the other answers are wrong. Choice A (1/4) represents the inverted ratio

\frac{A_2^2}{A_1^2}

, which would be the ratio

\frac{k_1}{k_2}

instead. Choice B (1/2) incorrectly uses the linear amplitude ratio

\frac{A_2}{A_1}

rather than the squared ratio. Choice C (2) similarly uses the linear ratio but in the wrong direction:

\frac{A_1}{A_2}

. Remember that energy in oscillatory systems depends on the square of the amplitude. When comparing systems with equal energy, the spring constant is inversely proportional to the square of the amplitude—smaller amplitudes require stiffer springs to store the same energy.

Question 13

A spring-mass system oscillates with amplitude A. At what fraction of the amplitude is the kinetic energy equal to 3 times the potential energy?

$\frac{1}{4}$
$\frac{1}{3}$
$\frac{1}{2}$ (correct answer)
$\frac{\sqrt{3}}{4}$
$\frac{3}{4}$

Explanation: When analyzing spring-mass systems, you need to understand how kinetic and potential energy trade off during oscillation. The total mechanical energy remains constant, but it's redistributed between these two forms as the mass moves. For a spring-mass system, the potential energy at position

x

U = \frac{1}{2}kx^2

, and the kinetic energy is

K = \frac{1}{2}mv^2

. The total energy equals the potential energy at maximum displacement:

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

. At any position

x

, we have

K + U = E

, so

K = E - U = \frac{1}{2}k(A^2 - x^2)

. Given that

K = 3U

, we can write:

K + U = 4U = E

, which means

U = \frac{E}{4}

. Since

U = \frac{1}{2}kx^2

and

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

, we have:

\frac{1}{2}kx^2 = \frac{1}{4} \cdot \frac{1}{2}kA^2

Simplifying:

x^2 = \frac{A^2}{4}

, so

x = \frac{A}{2}

The answer is C)

\frac{1}{2}

. Choice A)

\frac{1}{4}

would give you

K = 15U

, not

3U

. Choice B)

\frac{1}{3}

results from incorrectly thinking

U = \frac{E}{3}

when

K = 3U

. Choice D)

\frac{\sqrt{3}}{4}

comes from algebraic errors in solving the energy equation. Strategy tip: For energy problems in oscillators, always start with conservation of energy (

K + U = E

) and use the given ratio to express one energy in terms of the other. This systematic approach prevents algebraic mistakes.

Question 14

A mass-spring system oscillates with simple harmonic motion. When the displacement is half the amplitude, what fraction of the total energy is kinetic energy?

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$ (correct answer)
$\frac{\sqrt{3}}{2}$
$\frac{2}{3}$

Explanation: When analyzing energy in simple harmonic motion, remember that total mechanical energy remains constant and is split between kinetic energy (KE) and potential energy (PE). The key is understanding how this energy distribution changes with position. For a mass-spring system, the total energy is

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

, where

k

is the spring constant and

A

is the amplitude. At any displacement

x

, the potential energy is

PE = \frac{1}{2}kx^2

and kinetic energy is

KE = E - PE

. When the displacement is half the amplitude (

x = \frac{A}{2}

), the potential energy becomes:

PE = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{1}{2}k \cdot \frac{A^2}{4} = \frac{1}{4} \cdot \frac{1}{2}kA^2 = \frac{E}{4}

Therefore, the kinetic energy is:

KE = E - PE = E - \frac{E}{4} = \frac{3E}{4}

The fraction of total energy that is kinetic is

\frac{KE}{E} = \frac{3}{4}

, making C correct. Looking at the wrong answers: A)

\frac{1}{4}

represents the fraction that is potential energy at this position, not kinetic. B)

\frac{1}{2}

would be correct only at the specific displacement where

x = \frac{A}{\sqrt{2}}

. D)

\frac{\sqrt{3}}{2}

doesn't correspond to any standard energy fraction in SHM. Remember this pattern: in SHM, energy fractions follow the relationship

\frac{PE}{E} = \left(\frac{x}{A}\right)^2

. Square the displacement ratio to find the potential energy fraction, then subtract from 1 to get the kinetic fraction.

Question 15

Two identical pendulums oscillate with the same period but different amplitudes. Pendulum A has twice the amplitude of pendulum B. What is the ratio of their total energies Eₐ/Eᵦ?

1/4
1/2
2
4 (correct answer)
8

Explanation: When analyzing pendulum energy, you need to understand that a pendulum's total mechanical energy depends on its amplitude, not its period. For small oscillations, the total energy is entirely potential energy at maximum displacement. The total energy of a simple pendulum is

E = \frac{1}{2}k A^2

, where k is the effective spring constant and A is the amplitude. Since both pendulums are identical (same mass, length, and gravitational field), they have the same k value. This means energy is proportional to the square of the amplitude. Given that pendulum A has twice the amplitude of pendulum B (

A_A = 2A_B

), we can find the energy ratio:

\frac{E_A}{E_B} = \frac{\frac{1}{2}k(A_A)^2}{\frac{1}{2}k(A_B)^2} = \frac{(A_A)^2}{(A_B)^2} = \frac{(2A_B)^2}{(A_B)^2} = \frac{4A_B^2}{A_B^2} = 4

Therefore, the answer is D) 4. Choice A) 1/4 incorrectly inverts the relationship. Choice B) 1/2 might come from thinking energy is inversely proportional to amplitude. Choice C) 2 represents the common mistake of assuming energy is directly proportional to amplitude rather than to amplitude squared. Remember this key principle: for oscillatory motion, energy scales with the square of amplitude. Whether dealing with pendulums, springs, or other harmonic oscillators, doubling the amplitude always quadruples the total energy. This

A^2

relationship appears frequently in physics problems involving oscillations.

Question 16

A horizontal spring-mass system oscillates with total energy E₀. At the moment when the kinetic energy is E₀/4, what is the ratio of the restoring force magnitude to its maximum value?

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{\sqrt{3}}{2}$ (correct answer)
$\frac{3}{4}$
$\frac{\sqrt{2}}{2}$

Explanation: When you encounter spring-mass oscillation problems involving energy ratios, focus on the relationship between kinetic energy, potential energy, and position within the oscillation cycle. In simple harmonic motion, total energy

E_0

equals the sum of kinetic and potential energies at any instant. Since kinetic energy is

E_0/4

, the potential energy must be

3E_0/4

to conserve total energy. The potential energy in a spring system is

U = \frac{1}{2}kx^2

, where

k

is the spring constant and

x

is displacement from equilibrium. At maximum displacement (amplitude

A

), all energy is potential:

E_0 = \frac{1}{2}kA^2

. Since

U = 3E_0/4 = \frac{3}{4} \cdot \frac{1}{2}kA^2 = \frac{1}{2}kx^2

, we can solve for the current position:

x^2 = \frac{3A^2}{4}

, so

x = \frac{\sqrt{3}}{2}A

. The restoring force is

F = kx

, with maximum value

F_{max} = kA

at the amplitude. Therefore, the ratio is:

\frac{F}{F_{max}} = \frac{kx}{kA} = \frac{x}{A} = \frac{\sqrt{3}}{2}

Option A (

\frac{1}{4}

) incorrectly assumes the force ratio equals the kinetic energy fraction. Option B (

\frac{1}{2}

) might come from mistaking

\sqrt{1/4}

for the answer. Option D (

\frac{3}{4}

) incorrectly uses the potential energy fraction directly. Remember: in oscillation problems, always use energy conservation first to find position, then apply Hooke's law. The force is proportional to displacement, not directly to energy fractions.

Question 17

A mass attached to a vertical spring oscillates with amplitude 6.0 cm. The spring has natural length 20 cm and extends to 25 cm when the mass hangs at equilibrium. What is the minimum length of the spring during oscillation?

14 cm
19 cm (correct answer)
20 cm
21 cm
26 cm

Explanation: When analyzing vertical spring oscillations, you need to understand how the equilibrium position relates to the spring's natural length and how oscillation occurs around this equilibrium point. First, let's establish the equilibrium position. The spring's natural length is 20 cm, but it stretches to 25 cm when the mass hangs at rest. This means the equilibrium position is 5 cm below the natural length, where the spring force balances the gravitational force. During oscillation, the mass moves 6.0 cm above and below this equilibrium position. The key insight is that the minimum spring length occurs when the mass is at its highest point - 6.0 cm above equilibrium. At this point, the spring length is: 25 cm (equilibrium length) - 6.0 cm (amplitude above equilibrium) = 19 cm. Looking at the wrong answers: Choice A (14 cm) incorrectly assumes the spring compresses beyond its natural length, which would require the spring to push rather than just reduce its pull. Choice C (20 cm) represents the natural length, but the mass never reaches this high since it would require the spring force to completely overcome gravity plus provide additional upward acceleration. Choice D (21 cm) might result from incorrectly subtracting only 4 cm from equilibrium instead of the full amplitude. Remember that in vertical spring problems, always identify the equilibrium position first (where spring force equals weight), then apply the oscillation amplitude around that point. The spring will be shortest when the mass is highest in its motion.

Question 18

The graph shows the potential energy of a spring-mass oscillator as a function of position. Based on the graph, what is the kinetic energy when the mass is at position x = 0.05 m?

0.05 J
0.10 J
0.15 J (correct answer)
0.20 J
0.25 J

Explanation: From the graph, at x = 0.05 m, the potential energy U = 0.05 J. The maximum potential energy (total energy) occurs at the turning points where U = E = 0.20 J. Therefore, the kinetic energy at x = 0.05 m is K = E - U = 0.20 - 0.05 = 0.15 J. Choice A is the potential energy at this position. Choice B is half the total energy. Choice D is the total energy. Choice E assumes incorrect energy distribution.

Question 19

A mass-spring system oscillates with amplitude $A = 0.12$ m and angular frequency $\omega = 4.5$ rad/s. At time $t_1$ , the mass is at position $x = +0.06$ m and moving in the positive direction. What is the kinetic energy of the mass at this instant, expressed as a fraction of the total mechanical energy?

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$ (correct answer)
$\frac{\sqrt{3}}{2}$

Explanation: In SHM, the total energy is

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

, and at any position

x

, the potential energy is

U = \frac{1}{2}kx^2

. The kinetic energy is

K = E - U = \frac{1}{2}k(A^2 - x^2)

. Therefore,

\frac{K}{E} = \frac{A^2 - x^2}{A^2} = 1 - \frac{x^2}{A^2}

. With

x = 0.06

m and

A = 0.12

\frac{K}{E} = 1 - \frac{(0.06)^2}{(0.12)^2} = 1 - \frac{1}{4} = \frac{3}{4}

. Choice A gives the potential energy fraction. Choice B would be correct if

x = \frac{A}{\sqrt{2}}

. Choice D is not a valid energy ratio for this system.

Question 20

A horizontal mass-spring system has a total mechanical energy of $E = 0.25$ J. When the displacement from equilibrium is $x = +0.080$ m, the kinetic energy is $K = 0.16$ J. If the mass is $m = 0.50$ kg, what is the period of oscillation?

$0.89$ s
$1.26$ s (correct answer)
$1.57$ s
$2.51$ s

Explanation: First find the spring constant. The potential energy at

x = 0.080

m is

U = E - K = 0.25 - 0.16 = 0.09

J. Since

U = \frac{1}{2}kx^2

k = \frac{2U}{x^2} = \frac{2(0.09)}{(0.080)^2} = \frac{0.18}{0.0064} = 28.125

N/m. The period is

T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.50}{28.125}} = 2\pi\sqrt{0.0178} = 2\pi(0.133) = 1.26

s. Choice A uses

\omega = \sqrt{k/m}

instead of

T = 2\pi/\omega

. Choice C uses

k = 2E/x^2

incorrectly. Choice D uses

k = K/x^2

instead of

k = 2U/x^2