Question

A spring with k = 35.5 N/m has a mass of 5.50 kg attached to it. An external force F = (4.40 N)sin[(6.80 s−1)t] drives the spring mass system so that it oscillates without any resistive forces. What is the amplitude of the oscillatory motion of the spring-mass system?

Answer 1

Answer:

A = 0.02 m

Explanation:

The spring constant, k = 35.5 N/m

The attached mass, m = 5.50 kg

The expression for the external force, F =  (4.40 N)sin[(6.80 s⁻¹)t].....(1)

The general expression for the external force, F = F₀ sin (wt).............(2)

Comparing equations (1) and (2):

The forced frequency, $\omega = 6.80 rad/s$

F₀ = 4.40 N

The natural frequency can be calculated using the formula:

$\omega_0 = \sqrt{\frac{k}{m} } \\\\\omega_0 = \sqrt{\frac{35.5}{5.5} } \\\\\omega_0 = 2.54 rad/s$

The amplitude of oscillation of a spring-mass system in the steady state:

$A = \frac{F_0}{m(\omega^2 - \omega_o^2)} \\\\A = \frac{4.4}{5.5(6.8^2 - 2.54^2)} \\\\A = 0.02 m$