A particle executing SHM is described by the displacement function `x(t)=Acos(omegat+phi)`, if the initial (t=0) position of the particle is 1 cm, its initial velocity is `pii" cm "s^(-1)` and its angular frequency is `pis^(-1)`, then the amplitude of its motion is

Options

A. `picm`
B. `2cm`
C. `sqrt(2)cm`
D. `1cm`

Answer

A

Explanation

`x=Acos(omegat+phi)` where A is the amplitude.
At `t=0,x=1cmtherefore 1=Acosphi`
Velocity, `v=(dx)/(dt)=(d)/(dt)(Acos(omegat+phi))=-Aomegasin(omegat+phi)`
At `t=0,v=picms^(-1) therefore pi=-Aomegasinphior(pi)/(omega)=-Asinphi`
But `omega=pis^(-1)` (Given) `therefore 1=-Asinphi" "...(ii)`
Squaring and adding (i) and (ii), we get
`A^(2)cos^(2)phi+A^(2)sin^(2)phi=2orA=sqrt(2)cm(because sin^(2)phi+cos^(2)phi=1)`