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

Answer

B

Explanation

Let equation of simple harmonic motion is
`x=A sin(omega t+delta)`
It is given, A = 0.5 m and
`omega=(2pi)/(T)=(2pi)/(4)s^(-1)=(pi)/(2)s^(-1)`
At `t=0.5s,x=0.3m`
so `0.3=0.5sin(omega t+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[(pi t)/(2)-8^(@)]`