The displacement function of a `S.H.M` is given by
`y = cos [(omega t + phi)]`
if at `t = 0` th displaxcement is `y = 1` on and velocity `cm s^(-1)` The value amplitude `(A in cm)` is

Options

A. `1`
B. 1
C. `sqrt(2)`
D. `1//sqrt(2)`

Answer

C

Explanation

`y = A cos pi(t + phi) = A cos (pi t + pi phi)`
` v = Api sin(pi t+ pi phi)`
`At t = 0 , y_(0) = A cos pi phi rArr (y_(0))/(A) = cos pi phi`…(i)
`v_(0) = A pi sin pi phi rArr - (v_(0))/(Api) = sin pi phi`….(ii)
Squaring and adding we get `(y_(0)^(2))/(A^(2)) + (v_(0)^(2))/(A^(2)pi^(2)) = 1 rArr y_(0)^(2) + (gv_(0)^(2))/(pi^(2)) = A^(2)`
`y_(0) = 1 cm v_(0) = pi cms^(-1)`
`t^(2) + (pi^(2))/(pi^(2)) = A^(2) or A = sqrt(2) cm`