The path length of oscillation of simple pendulum of length 1 m is 16 cm. Its maximum velocity is (take, `g=pi^(2)m//s^(2)`)

Options

A. `2picm//s`
B. `4picm//s`
C. `8picm//s`
D. `16picm//s`

Answer

C

Explanation

Given,
Length of the pendulum `(l)=1m`
`:.` Amplitude `(a)=("Path length")/(2)=(16)/(2)=8cm`
Acceleration due to gravity `(g)=pi^(2)m//s^(2)`
We know that, time period `(T)=2pisqrt((l)/(g))=2pisqrt((1)/(pi^(2)))`
`T=(2pi)/(pi)`
`T=2s`
`:.` Maximum velocity `(v_(max))=aomega`
`=axx(2pi)/(T)" "(becauseomega=(2pi)/(T))`
`=(8xx2xxpi)/(2)8picm//s`