The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is `10 s^(-1)` At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is `(pi)/(4)`

Options

A. 5`500m/s^(2)`
B. `750sqrt2m/s^(2)`
C. `700m/s^(2)`
D. `500sqrt2m/s^(2)`

Answer

D

Explanation

`a_(max) = omega^(2)A`
`v_(max) = omegaA`
`(a_(max))/(v_(max)) = 10 s^(-1) " "rArr (omega^(2)A)/(omegaA) = 10`
`rArr omega = 10 rad//s`
Let `x = A cos (omegat + phi)`
At ` t = 0 " " 5 = A cos [(omega + 0) + (pi)/(4)]`
` 5 = A cos((pi)/(4))`
`5 = A ((1)/(sqrt(2))) " " rArr A = 5sqrt(2)`
Maximum acceleration,
`a_(max) = omega^(2)A = 500sqrt(2) ms//s^(2)`