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: Understand the concept
In simple harmonic motion the speed is greatest at the mean (equilibrium) position and equals \(v_{max} = A\omega\).

Step 2: Find \(\omega\)
For a spring-mass system \(\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{180}{0.2}} = \sqrt{900} = 30\) rad/s.

Step 3: Compute
The amplitude is \(A = 4\ \text{cm} = 0.04\) m, so
\[ v_{max} = A\omega = 0.04\times30 = 1.2\ \text{m/s} \]

Step 4: Check
The result is option (C). The amplitude must be in metres. Using \(A = 4\) with \(\omega = 30\) would give 120, and a slip of a factor of 10 in the amplitude would give 0.12 or 12, which are the other listed values.

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