Answer:
[tex]velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)[/tex]
Max speed = [tex]\frac{15\, \pi}{4} \,\, \frac{m}{s}[/tex]
Max acceleration = [tex]\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}[/tex]
Explanation:
Given the description of period and amplitude, the SHM could be described by:
[tex]f(x)=5\,sin(\frac{\pi}{4}x)[/tex]
and its angular velocity can be calculated doing the derivative:
[tex]f(x)=5\, \,sin(\frac{\pi}{4}x)\\f'(x)=5\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)[/tex]
And therefore, the tangential velocity is calculated by multiplying this expression times the radius of the movement (3 m):
[tex]velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)[/tex] and is given in m/s.
Then the maximum speed is obtained when the cosine function becomes "1", and that gives:
Max speed = [tex]\frac{15\, \pi}{4} \,\, \frac{m}{s}[/tex]
The acceleration is found from the derivative of the velocity expression, and therefore given by:
[tex]acceleraton(x)=-15\,\frac{\pi^2}{16}\,sin(\frac{\pi}{4}x)[/tex]
and the maximum of the function will be obtained when the sine expression becomes "-1", which will render:
Max acceleration = [tex]\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}[/tex]
