The position velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2cm, 1ms^-1 and 10ms^-2 at a certain instant. Find the amplitude and the time period of the motion.

Answer

B, C, D

Explanation

Given that at a particular instant
`x=2cm=0.02m,
`v=1m sec^-1`
`a=10 m sec^-2`
We know that
`a=w^2x`
`rarr omega=sqrta/xsqrt10/0.02`
`=sqrt500=10sqrt5`
`T=(2pi)/omega=(2pi)/(10sqrt5)`
`=(2xx3.14)/(10xx2.236)`
`=.28second
Again amplitude r is given by
`v=wsqrt(r^2-x^2)`
`rarr v^2=w^2(r^2-x^2)`
`1=500(r^2-0.0004)`
`rarr r=0.0489=0.049m`
`rarr r=4.9cm`