Question:
A spring of force constant \(180\) N/m is loaded with a mass \(0.20\) kg. The amplitude of oscillations is \(4\) cm. When mass comes to equillibrium position, its velocity in m/s is
A spring of force constant \(180\) N/m is loaded with a mass \(0.20\) kg. The amplitude of oscillations is \(4\) cm. When mass comes to equillibrium position, its velocity in m/s is
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At the mean position the speed is maximum and equals amplitude times angular frequency.
Updated On: Nov 12, 2026
- \(12\)
- \(1.2\)
- \(0.12\)
- \(0.012\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
At the equilibrium position the speed is maximum: \(v=A\omega\).
Step 2: Angular frequency
\(\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{180}{0.20}}=\sqrt{900}=30\) rad/s.
Step 3: Calculation
\(v=0.04\times30=1.2\) m/s.
Final Answer:
The velocity at the equilibrium position is 1.2 m/s. \[ \boxed{\text{(B) }1.2} \]
At the equilibrium position the speed is maximum: \(v=A\omega\).
Step 2: Angular frequency
\(\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{180}{0.20}}=\sqrt{900}=30\) rad/s.
Step 3: Calculation
\(v=0.04\times30=1.2\) m/s.
Final Answer:
The velocity at the equilibrium position is 1.2 m/s. \[ \boxed{\text{(B) }1.2} \]
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