Answer to Question #99497 in Mechanics | Relativity for Aled

Question

A machine is mounted on springs. When operating, the base of the machine is found to be vibrating up and down with a period of 271.5 ms. The displacement of the machine base from its equilibrium height, h, varies as:

h(t) = A sin (ωt)

where A = 0.22 mm is the amplitude of the vibration, and ω is the angular frequency of the vibration.

What is the height of the machine above its equilibrium position at a time, t = 5.7 s?

Give your answer in mm to two decimal places. Note that the height relative could be positive or negative and you have enter the negative sign if it is negative.

Answer 1

Displacement function or height function is given as

$$h(t)=Asin(\omega t)$$

$\omega=2\pi/T$

T=271.5ms=0.2715s

thus

ω=2x3.14/0.2715= 23.13076 per second

A=0.22mm

Thus

h at t=5.7 second =0.22sin (23.13076x5.7) = -0.0223mm