A simple harmonic motion along the x-axis has the following properties: amplitude `=0.4m` the time to go from one extreme position to other is `2s` and `x=0.3m` at `t=0.5`a. the general equation of the simple harmonic motion is

Options

A. `x=(0.5m)sin[(pit)/(2)+8^@]`
B. `x=(0.5m)sin[(pit)/(2)-8^@]`
C. `x=(0.5m)cos[(pit)/(2)+8^@]`
D. `x=(0.5m)cos[(pit)/(2)-8^@]`

Answer

B

Explanation

Let equation of simple harmonic motion is
`x=Asin(omegat+delta)`
it is given, `A=0.5m` and `omega=(2pi)/(T)=(2pi)/(4)s^-1=(pi)/(2)s^-1`
At `t=0.5s`,`x=0.3m`, so `0.3=0.5sin(omegat+delta)`
`impliessin((pi)/(2)xx(1)/(2)+delta)=(3)/(5)implies(pi)/(4)+delta=37^@`
`impliesdelta=37^@-45^@=-8^@`
So, equation of motion is `x=(0.5m)sin[(pit)/(2)-8^@]`