Question

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

Answer 1

Answer:

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$

Explanation:

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

$f(x)=5\,sin(\frac{\pi}{4}x)$

and its angular velocity can be calculated doing the derivative:

$f(x)=5\, \,sin(\frac{\pi}{4}x)\\f'(x)=5\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

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

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$  and is given in m/s.

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

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

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

$acceleraton(x)=-15\,\frac{\pi^2}{16}\,sin(\frac{\pi}{4}x)$

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

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$