Question

A particle is moving with (SHM) of period 8.0s and amplitude5.0m

Find
The speed of the particle when it is 3.0m from the centre of it's motion,
The maximum speed,
The maximum acceleration

Answer 1

`velocity(x)=15\,(\pi)/(4)\,cos((\pi)/(4)x)`

Max speed =
`(15\, \pi)/(4) \,\, (m)/(s)`

Max acceleration =
`(15\,\pi^2)/(16) \,\,(m)/(s^2)`

Step-by-step explanation:

Given the description of period and amplitude, the SHM could be described by:

`f(x)=5\,sin((\pi)/(4)x)`

and its angular velocity can be calculated doing the derivative:

`f(x)=5\, \,sin((\pi)/(4)x)\\f'(x)=5\,(\pi)/(4)\,cos((\pi)/(4)x)`

And therefore, the tangential velocity is calculated by multiplying this expression times the radius of the movement (3 m):

`velocity(x)=15\,(\pi)/(4)\,cos((\pi)/(4)x)` and is given in m/s.

Then the maximum speed is obtained when the cosine function becomes "1", and that gives:

Max speed =
`(15\, \pi)/(4) \,\, (m)/(s)`

The acceleration is found from the derivative of the velocity expression, and therefore given by:

`acceleraton(x)=-15\,(\pi^2)/(16)\,sin((\pi)/(4)x)`

and the maximum of the function will be obtained when the sine expression becomes "-1", which will render:

Max acceleration =
`(15\,\pi^2)/(16) \,\,(m)/(s^2)`