A particle executes `SHM` with an amplitude of `10cm` and frequency `2 Hz`. At `t = 0`, the particle is at a point where potential energy and kinetic energy are same. The equation for its displacement is

Options

A. `0.1 sin(4pi t+(pi)/(4))`
B. `0.1 sin 4pi t`
C. `0.1 cos(4pi t+(pi)/(4))`
D. None of above

Answer

A

Explanation

Let x= A sin `(omegat+phi)`
`therefore" "v=(dx)/(dt)=A omega cos (omegat+phi)`
`therefore" "KE=(1)/(2)mv^(2)=(1)/(2)mA^(2)omega^(2)cos^(2)(omegat+phi)`
`therefore" "(KE)_("max")=(1)/(2)mA^(2)omega^(2)`
`therefore" "PE=(1)/(2)mA^(2)omega^(2)-KE`
`=(1)/(2)mA^(2)omega^(2)-(1)/(2)mA^(2)omega^(2)cos^(2)(omegat+phi)`
`=(1)/(2)mA^(2)omega^(2)sin^(2)(omegat+phi)`
According to problem, `KE=PE" "("at t"=0)`
`therefore" "=(1)/(2)mA^(2)omega^(2)sin^(2)(omegat+phi)`
`=(1)/(2)mA^(2)omega^(2)cos^(2)(omegat+phi)`
`therefore tan^(2)(omegat+phi)=1`
`rArr tan^(2)(omegat+phi)=tan^(2).(pi)/(4)`
`rArr" "omegat+phi=(pi)/(4)`
`rArr" "phi=(pi)/(4)" "(becauset=0)`
`because" "x=A sin(omegat+phi)`
Here, A = 10 cm =`0.1`m
`omega=2pi f=2pixx2=4pi"rad"//s`
`phi =(pi)/(4)`
`therefore" "x=0.1 sin (4pi t+(pi)/(4))`