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A particle executing SHM is described by...

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

A

`picm`

B

2 cm

C

`sqrt(2)`cm

D

1 cm

Text Solution

Verified by Experts

The correct Answer is:
C
`x=Acos(omegat-phi)` where A is amplitude.
At t=0, x=1 cm
`therefore1=Acosphi` . . . (i)
Velocity, `v=(dx)/(dt)=(d)/(dt)(aCos(omegat+phi))=-Aomegasin(omegat+phi)` . . . (ii)
Squaring and adding (i) and (ii), we get `A^(2)cos^(2)phi+A^(2)sin^(2)phi=2`
`A^(2)=2" "(becausesin^(2)phi+cos^(2)phi=1)`
`thereforeA=sqrt(2)cm`
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