A clock pendulum oscillates at a frequency of 2.5 Hz . At t=0, it is released fr

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.

Part A Ignoring friction, what will be the position (angle in radians) of the pendulum at t = 0.25 s ? Express your answer using two significant figures.

Part B Ignoring friction, what will be the position (angle in radians) of the pendulum at t = 2.00 s ? Express your answer using two significant figures.

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.

Explanation / Answer

Solution: From the question we have

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

So, if you visualize the pendulum, it starts at 15º. Then, it swings to the other side. It stops at the exact same height on the other side (at -13º, on the other side), and swings back and forth. It goes back and forth between the two positions.

From this, we know that the amplitude of the swing is at 13º, since that is the maximum. We also know that its frequency is 2.5. So, putting this together, the equation for its angle at time t is,

angle = 15º * cos(360º *2.5 * t)

Part A Ignoring friction, what will be the position (angle in radians) of the pendulum at t = 0.25 s ?

angle = 15º * cos(360º *2.5 * t)

=>angle=15º * cos(360º *2.5 * 0.25) =>angle= (-13.86) degree

Part B Ignoring friction, what will be the position (angle in radians) of the pendulum at t = 2.00 s ?

angle = 15º * cos(360º *2.5 * t)

=>angle= 15º * cos(360º *2.5 * 2.0)

=>angle=(-15) degree

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

=>angle= 15º * cos(360º *2.5 * 540.0)

=>angle=15 degree

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