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, [tex]\omega = 6.80 rad/s[/tex]
F₀ = 4.40 N
The natural frequency can be calculated using the formula:
[tex]\omega_0 = \sqrt{\frac{k}{m} } \\\\\omega_0 = \sqrt{\frac{35.5}{5.5} } \\\\\omega_0 = 2.54 rad/s[/tex]
The amplitude of oscillation of a spring-mass system in the steady state:
[tex]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[/tex]
