Physics: Simple Harmonic Motion and Pendulums Worksheet

Question 1

A 0.50 kg mass is attached to a spring with spring constant 200 N/m. Calculate the period of the mass-spring oscillator.

Accepted Answer

The period is about 0.314 s. Using T = 2π sqrt(m/k), T = 2π sqrt(0.50/200) = 2π sqrt(0.0025) = 0.314 s.

Question 2

A simple pendulum has a length of 1.00 m. Calculate its period and frequency for small oscillations.

Accepted Answer

The period is about 2.01 s, and the frequency is about 0.50 Hz. Using T = 2π sqrt(L/g), T = 2π sqrt(1.00/9.8) = 2.01 s, and f = 1/T = 0.50 Hz.

Question 3

A pendulum has a period of 3.0 s. Calculate the length of the pendulum.

Accepted Answer

The length is about 2.23 m. Rearranging T = 2π sqrt(L/g) gives L = g(T/2π)^2, so L = 9.8(3.0/2π)^2 = 2.23 m.

Question 4

A 0.25 kg mass on a spring completes one full cycle every 0.50 s. Calculate the spring constant.

Accepted Answer

The spring constant is about 39.5 N/m. Using T = 2π sqrt(m/k) and rearranging gives k = 4π^2m/T^2, so k = 4π^2(0.25)/(0.50)^2 = 39.5 N/m.

Question 5

A spring with spring constant 80 N/m oscillates with an amplitude of 0.10 m. Find the total mechanical energy of the oscillator. Then find the spring potential energy and kinetic energy when the displacement is 0.06 m.

Accepted Answer

The total mechanical energy is 0.40 J. At x = 0.06 m, the spring potential energy is 0.144 J, and the kinetic energy is 0.256 J because total energy is conserved.

Question 6

A mass-spring oscillator has an amplitude of 0.15 m and a period of 1.20 s. Calculate the maximum speed and maximum acceleration of the mass.

Accepted Answer

The maximum speed is about 0.79 m/s, and the maximum acceleration is about 4.11 m/s^2. The angular frequency is ω = 2π/T = 5.24 rad/s, so vmax = Aω = 0.15(5.24) and amax = Aω^2 = 0.15(5.24)^2.

Question 7

For a mass on a spring moving in simple harmonic motion, describe the speed, acceleration, and force when the mass is at the equilibrium position.

Accepted Answer

At the equilibrium position, the speed is maximum, while the acceleration and net force are zero. The spring is not stretched or compressed at that instant, so the restoring force is zero.

Question 8

A pendulum is moved from Earth to a planet where the acceleration due to gravity is smaller. Explain what happens to the pendulum's period if its length stays the same.

Accepted Answer

The pendulum's period increases. Since T = 2π sqrt(L/g), a smaller value of g makes the square root larger, so each swing takes more time.

Question 9

A student counts 15 complete oscillations of a mass-spring system in 30.0 s. Calculate the period and frequency.

Accepted Answer

The period is 2.0 s, and the frequency is 0.50 Hz. The period is total time divided by cycles, 30.0 s divided by 15, and the frequency is 1/2.0 s.

Question 10

A 0.20 kg mass attached to a spring with spring constant 50 N/m is displaced 0.040 m to the right of equilibrium. Taking right as positive, calculate the acceleration at that instant.

Accepted Answer

The acceleration is -10 m/s^2. Using F = -kx and a = F/m gives a = -(k/m)x = -(50/0.20)(0.040) = -10 m/s^2, so the acceleration is to the left toward equilibrium.

Question 11

The position of an oscillator is modeled by x(t) = A cos(ωt). Its amplitude is 0.050 m and its period is 0.80 s. If the object starts at maximum positive displacement, find its position at t = 0.20 s and at t = 0.40 s.

Accepted Answer

At t = 0.20 s, the position is 0 m because this is one-fourth of a period after the start. At t = 0.40 s, the position is -0.050 m because this is one-half of a period after the start.

Question 12

A pendulum of length 0.75 m has a measured period of 1.74 s. Use the data to calculate the experimental value of g.

Accepted Answer

The experimental value of g is about 9.78 m/s^2. Rearranging T = 2π sqrt(L/g) gives g = 4π^2L/T^2, so g = 4π^2(0.75)/(1.74)^2 = 9.78 m/s^2.

Physics: Simple Harmonic Motion and Pendulums

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