Question Detail
Question
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately
Correct Answer
Option D is the correct answer.
Detailed Explanation
The spring stretches 10 cm (0.10 m) under a 5.0 kg mass, so k = mg/x = (5.0)(9.8)/0.10 = 490 N/m. The angular frequency ω = √(k/m) = √(490/5.0) = √98 ≈ 9.9 rad/s. The mass is displaced an additional 5.0 cm (0.05 m) from equilibrium and released from rest. This means the amplitude A = 0.05 m, and the initial condition y(0) = A means the cosine function is used. Therefore, y = 0.05 cos 9.9t.
Hint
Find k from the static stretch using F = kx. Then find ω = √(k/m). The amplitude is the additional displacement, and since it starts at maximum displacement, use cosine.