Question

If the period of oscillation of a simple pendulum is 4s, find its length. If the velocity of the bob

at the mean position is 40cms−1
, find its amplitude. (take gravity = 9.81ms−2

Answer 1

Step-by-step explanation:

Because we assume the pendulum is a "mathematical pendulum" (neglecting the moment of inertia of the bob), we can find:

`T=2\pi\sqrt{(L)/(g)} \rightarrow 4=2\pi\sqrt{(L)/(9.81)} \rightarrow (4)/(\pi^(2))=(L)/(9.81) \rightarrow L \approx 3.97 m`

By using the
`y=A\sin(\omega t) \rightarrow v = (dy)/(dt)=\omega A \cos\omega t = \omega\sqrt{A^(2)-y^(2)}`

The mean position is the position when y = 0, so:

`\omega = (2\pi)/(T)=(2\pi)/(4)=0.5\pi` rad/s

and
`v = \omega A \rightarrow A=(40)/(0.5\pi)=(80)/(\pi)` in centimeters (cm).