Question

A body vibrating with a simple harmonic motion has a frequency of 4hz and Amplitude of 0.15m. Calculate the maximum value of acceleration and velocity.

b) the acceleration and velocity at a point 0.09 from the equilibrium position of motion

Answer 1

Solution:

$a_{max} ​ =(2πf) ^2 A=(2(4)π)^2 0.15=95(\tfrac{m}{s^2})$

$v_{max} ​ =(2πf)A=(2(4)π)0.15=3.8(\tfrac{m}{s})$

For the second part:

$a=(2(4)π) ^2 0.09=57(\tfrac{m}{s^2})$

$v=(2(4)π)0.09=2.3(\tfrac{m}{s})$