Question

A spring force constant \(180\) N/m is loaded with a mass \(0.2\) kg. The amplitude of oscillations is \(4\) cm. When mass comes to equilibrium position, its velocity is

• A. \(0.012\) m/s
• B. \(0.12\) m/s
• C. \(1.2\) m/s
• D. \(12\) m/s

Hint:

The speed at the equilibrium position is the maximum speed, A times omega.

Answer 1

The Correct Option is C

Step 1: Energy method
At the extreme position all energy is potential: $\frac{1}{2}kA^2$. At equilibrium it is all kinetic: $\frac{1}{2}mv^2$.

Step 2: Equate
$v = A\sqrt{\dfrac{k}{m}} = 0.04\times\sqrt{900}$.

Step 3: Compute
$v = 0.04\times30 = 1.2$ m/s.

Step 4: Check
Energy: $\frac{1}{2}(180)(0.04)^2 = 0.144$ J and $\frac{1}{2}(0.2)(1.2)^2 = 0.144$ J. They match.

Final Answer:
The speed at equilibrium is 1.2 m/s. This is option (C). \[ \boxed{\text{(C) }1.2\ \text{m/s}} \]