Question

Find the time period of the S.H.M described by `x=3*sin(pi/2t)`?

Answer 1

The period is `=4s`

Explanation:

The displacement of SHM is

`x=asin(omegat+phi)`

Comparing this to the equation

`x=3sin(1/2pit)`

The amplitude is `a=3`

The angular frequency is

`omega=1/2pi`

Therefore,

The period is

`T=(2pi)/omega=(2pi)/(1/2pi)=4s`

Answer 2

The period is 4 s.

Explanation:

The period will be the time for `sin(pi/2*t)` to increase from `sin0 " to" sin(2pi)`. That is because `sin(2pi)` will be the first time that the wave is at the same amplitude as when it started at sin0. That amplitude was at the midpoint between max crest and max trough and also on the upswing. The value of the function at those 2 times will be zero as demonstrated in the following paragraph.

The following graph plots the value of t versus the value of `3sin(pi/2*t)` so that, as an example, at 2 on the x (actually, the t) axis the value of `pi/2*t " is " pi` and the value of the y axis is zero `-` on the downswing. `sinpi = 0`, so that is as it should be.
graph{y = 3sin((pi/2)x) [-0.15,5.5,-5,5]}

So the point on the time axis when the plot is repeating what it did at t=0 is at t=4 s. Therefore the period is 4 s.

I hope this helps,
Steve