Question

A particle executies linear simple harmonic motion with an amplitude `3cm` .When the particle is at `2cm` from the mean position , the magnitude of its velocity is equal to that of acceleration .The its time period in seconds is
A. `sqrt(5)/(2 pi)`
B. `(4pi)/(sqrt(5))`
C. `(2pi)/(sqrt(3))`
D. `sqrt(5)/(pi)`

Answer 1

Correct Answer - B
Amplitude `A = 3 cm`
When particle is at `x = 2 cm`
its `|"velocity"| = |"acceleration"|`
i.e. `omega sqrt(A^(2) - x^(2)) = omega^(2)x rArr omega = sqrt(A^(2) - x^(2))/(x)`
`T = (2pi)/(omega) = 2pi ((2)/(sqrt(5))) = (4pi)/(sqrt(5))`