A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6s. At t=0 it is at position x=5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t=4s.

Answer

A, B, C

Explanation

Given `r=10cm, At t=0, n=5cm T=6sec`
So, `omega=(2pi)/T=(2pi)/6=pi/3 sec^-1`
Att=0, x=5cm
so, `5=10sin(wx0+phi)`
`[Y=rsinwt] `
`=10sinphi`
sinphi=1/2`
`rarr phi=pi/6`
`:.` Equation of displacement
`=x=(10cm)sin(pi/3t+pi/6)`
ii. At t=4 second
`x=10sin[pi/34+pi/6]`
`=[10sin[(8pi+pi)/6)]`
`=10sin((9pi)/6)`
`=10sin((3pi)/2)`
`=10sinpi+pi/2`
`=-10sinpi/2=-10`
Acceleration
`a=-w^2x`
`=((-pi^2)/9)xx(-10)`
`=10.9~~0.11cm/sec`