A mass of 03kg is suspended from a spring of stiffness 200Nm

A mass of 0.3kg is suspended from a spring of stiffness 200Nm^-1. If the mass is displaced by 10mm from its equilibrium position and released, for the resulting vibration, calculate: A) (i) the frequency of the vibration (ii) the maximum velocity of the mass during the vibration (iii) the maximum acceleration of the mass during the vibration (iv) the mass required to produce double the maximum velocity calculated in (ii) using the same spring and initial deflection. B) plot a graph of acceleration against displacement (x)(for values of x from x=-10mm to x=+10mm)

Solution

solution

1)

F= 1 / (2) x (200/0.3) = 4.11Hz

Then I found the Angular Freq using = (200/0.3) = 25.82 rads -1

A = diplacement
= Angular Freq
k = Spring Stiffness

2)

Vmax = A x = 0.258 m/s-1

3)

Amax = ^2 x A = 6.67 m/s-2

4)

Mass required:

M = k / ^2 = 200 / 51.63977795^2 = 0.075 kg = 750e-3 Kg