Question

For the simple harmonic motion equation d = 5sin(2pi t), what is the maximum displacement from the equilibrium position?

Answer 1

The max displacement is the amplitude that is 5

Answer 2

d=
`5`

Explanation:

The given simple harmonic motion equation is given as:

d=
`5sin(2{\pi}t)`

Differentiating the above equation with respect to t, we have

`d'=5cos(2{\pi}t)(2{\pi)`

`d'=10{\pi}cos(2{\pi}t)`

Now, equating
`d'=0`, we have

`10{\pi}cos2{\pi}t=0`

`cos2{\pi}t=0`

`2{\pi}t=\frac{{\pi}}{2}`

`t=(1)/(4)`

Also, differentiating again with respect to t, we get

`d''=10{\pi}(-sin(2{\pi}t)(2{\pi})`

⇒
`d''<0`

Therefore, at t=
`(1)/(4)`, d=
`5sin(2{\pi}t)` will have maximum displacement.

⇒d=
`5sin(2{\pi}((1)/(4)))`

⇒d=
`5sin((\pi)/(2))`

⇒d=
`5`