Question Detail
Question
A 2-kg object stretches a spring 10 cm when it hangs vertically in equilibrium. If this object is then attached to the same spring on a smooth table and released from rest, what is the angular frequency of the resulting simple harmonic motion? (g = 9.81 m/s²)
Correct Answer
Option B is the correct answer.
Detailed Explanation
From Worked Example 16 on page 32: When the object hangs vertically in equilibrium, the spring force balances gravity: F_s = mg. So, kΔL = mg. The spring constant k = mg/ΔL = (2 kg * 9.81 m/s²) / 0.10 m = 196.2 N/m. The angular frequency ω = √(k/m) = √(196.2 N/m / 2 kg) = √98.1 ≈ 9.90 rad/s.
Hint
First, use the vertical equilibrium condition to find the spring constant 'k'. Then, use 'k' and the mass 'm' to find the angular frequency for horizontal SHM.