Question

A body executes `SHM`, such that its velocity at the mean position is `1ms^(-1)` and acceleration at exterme position is `1.57 ms^(-2)`. Calculate the amplitude and the time period of oscillation.

Answer 1

`(a_(max))/(v_(max)) = (A•^(2))/(A•) = (1.57)/(1) rArr• = 1.57 rad`
`:.` Time period `T = (2•)/(1.57) = (2(3.14))/(1.57) = 4s`.
but `A• = 1`, i.e., `A (1.57) = 1 or A = (1)/(1.57)`
`:.` Amplitude `A = 0.637m`.