A particle starts simple harmonic motion from the mean position. Its amplitude is a and time period is T. what is its displacement when its speed is half of its displacement when its speed is half of its maximum speed?

Options

A. `2a`
B. `(a)/(2) sqrt(3)`
C. `a`
D. `(a)/(2)`

Answer

B

Explanation

Using the relation for velocity, `v = omega sqrt((a^(2)-y^(2)))`
and `v_(max) = a omega and v'=(v_(max))/(2)`
So, `omega sqrt(a^(2)-y^(2))=(omega a)/(2)=a^(2)-y^(2)=(a^(2))/(4)`
So, `y^(2)=(3)/(4)a^(2) rArr y = (a sqrt(3))/(2)`