Question

A particle executes simple harmonic oscilliation with an amplitude `a`. The period of oscillations is `T`. The minimum time taken by the particle to travel half to the amplitude from the equlibrium position is
A. `(T)/(4)`
B. `(T)/(8)`
C. `(T)/(12)`
D. `(T)/(2)`

Answer 1

Correct Answer - C
Let diaplacement equation of particle excuting SHM is
`y= a sin omega t`
Therefore, `(a)/(2)= a sin omega t`
or `sin omega t=(1)/(2)= (sin)(pi)/(6)`
or `omega t= (pi)/(6)`
or `t=(pi)/(6 omega)`
or `t=(pi)/(6(2pi/T)) (as omega= (2pi)/(T))`
or `t=(T)/(12)`
Hence, the particle travels half to the amplitude from the equilibrium in `(T)/(12)s`.