A particle executes S.H.M. of amplitude 25 cm and time period 3 s. What is the minimum time required for the particle to move between two points 12.5 cm on either side of the mean position ?

Options

A. 0.6 s
B. 0.5 s
C. 0.4 s
D. 0.2 s

Answer

B

Explanation

Here, amplitude of particle, A = 25 cm
and time period, T = 3 s
If the particle at t = 0 s is at mean position, its displacement equation will be
`x = A sin omega t`
i.e., `x = 25 "sin"(2 pi t)/(3)" "[because omega = 2 pi v = (2pi)/(T)=(2pi)/(3)]`
If it takes time `t_(1)` to move a distance x = 12.5 cm to one side of its mean position, then `12.5 = 25"sin"(2 pi t_(1))/(3)or (1)/(2)="sin"(2 pi t_(1))/(3)`
or `"sin"(pi)/(6)="sin"(2 pi t_(1))/(3)`
`therefore" "(pi)/(6)=(2 pi t_(1))/(3)rArr t_(1)=(1)/(4)`
The same will be the time to move 12.5 cm to the other side of its mean position therefore, total time
`t = t_(1)+t_(2) rArr =(1)/(4)+(1)/(4)=(1)/(2)=0.5s`