A clock pendulum oscillates at a frequency of 25 Hz At t0 i

A clock pendulum oscillates at a frequency of 2.5 Hz . At t=0, it is released from rest starting at an angle of 16 to the vertical. Ignoring friction, what will be the position (angle in radians) of the pendulum at t = 2.80 s ?

Solution

For a pendulum the general equation for its motion is given as:

= A cos(t) where is the displacement of the pendulum in radians, is the angular frequency and t is the time in seconds.

Now, for frequency f, we know that the time period is given as T = 1/f

Also, T = 2/

or, = 2f = 2 x 3.14 x 2.5 = 15.7 rad/s

Also, the pendulum has been released from rest at an angle of 16 degrees which implies that the amplitude of the pendulum is 16 degrees = 0.279 radians

Hence we have the equation of motion for the pendulum as: = (0.279) cos(15.7t)

Now, we need to determine the value of angular displacement at t = 2.8 seconds, we can substitute the value t = 2.8 seconds in the equation above to obtain:

= (0.279) cos(15.7 x 2.8) = 0.2008 radians from the vertical.