1.

A particle executes simple harmonic motion with a period of `16s`. At time `t=2s`, the particle crosses the mean position while at `t=4s`, its velocity is `4ms^-1` amplitude of motion in metre isA. `sqrt(2)pi`B. `16sqrt(2)pi`C. `32sqrt(2)//pi`D. `4//pi`

Answer» Correct Answer - C
`y_(1) = A sin ((2pi)/(T)xxt), V = omega sqrt(A^(2) - y_(1)^(2))`


Discussion

No Comment Found

Related InterviewSolutions