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 11 ∘ to the vertical.

Part A Ignoring friction, what will be the position (angle in radians) of the pendulum at t = 0.35 s ? Express your answer using two significant figures. θ = rad

Part B Ignoring friction, what will be the position (angle in radians) of the pendulum at t = 1.60 s ? Express your answer using two significant figures. θ = rad

Part C Ignoring friction, what will be the position (angle in radians) of the pendulum at t = 540 s ? Express your answer using two significant figures. θ =

Answer 1

Solution:

Given:

Frequency (f) = 2.5 Hz .

At t=0, it is released from rest starting at an angle of 11^∘ to the vertical.

Thus: the amplitude of the swing is at 11º {since that is the maximum}.

The equation for its angle at time t is,

Angle () = 11º * cos(360º *2.5 * t)

Part A

Ignoring friction, what will be the position (angle in radians) of the pendulum at t = 0.35 s ?

Angle () = 11º * cos(360º *2.5 * 0.35)

Angle () = 7.78^o = 0.136 rad

Part B

Ignoring friction, what will be the position (angle in radians) of the pendulum at t = 1.60 s ?

Angle () = 11º * cos(360º *2.5 * 1.60)

Angle () = 11º = 0.192 rad

Part C

Ignoring friction, what will be the position (angle in radians) of the pendulum at t = 540 s ?

Angle () = 11º * cos(360º *2.5 * 540)

Angle () = 11º = 0.192 rad