Question

A clock pendulum oscillates at a frequency of 2.5 Hz . At t=0, it is released from rest starting at an angle of 13 ∘ to the vertical.a. What will be the position (angle in radians) of the pendulum at t = 0.25 s ? Express your answer using two significant figures.b. What will be the position (angle in radians) of the pendulum at t = 2.00 s ? Express your answer using two significant figures.

Answer 1

The pendulum's position at t = 0.25 s will be 0 radians (equilibrium position), and at t = 2.00 s, it will return to its maximum displacement of 0.227 radians from the vertical.

Step-by-step explanation:

When calculating the position of a pendulum in simple harmonic motion, we can use the formula θ(t) = θmax·cos(2π·f·t), where θmax is the maximum angle in radians, f is the frequency of oscillation, and t is the time elapsed. Given that the frequency f is 2.5 Hz and the time is 0.25 s:

The maximum angle in radians is θmax = 13° · (π/180) = 0.227 radians.

θ(0.25 s) = 0.227 · cos(2·π·(2.5)·(0.25)) = 0.227 · cos(π/2) = 0.

At 0.25 seconds, the pendulum will be at the equilibrium position, which is 0 radians from the vertical.

For t = 2.00 s:

θ(2.00 s) = 0.227 · cos(2·π·(2.5)·(2.00)) = 0.227 · cos(10π) = 0.227 radians, since cos(10π) = 1

At 2.00 seconds, the pendulum will be at its maximum displacement, which is 0.227 radians from the vertical.