Question:
A body of mass $1~\text{kg$ is attached to the lower end of a vertically suspended spring of force constant $600~\text{N/m}$. If another body of mass $0.5~\text{kg}$ moving vertically upward hits the suspended body with a velocity $3~\text{m/s}$ and is embedded in it, then the frequency of the oscillation is}
A body of mass $1~\text{kg$ is attached to the lower end of a vertically suspended spring of force constant $600~\text{N/m}$. If another body of mass $0.5~\text{kg}$ moving vertically upward hits the suspended body with a velocity $3~\text{m/s}$ and is embedded in it, then the frequency of the oscillation is}
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For spring oscillations after mass attachment, use $f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}$ with the total mass.
Updated On: Jun 4, 2025
- $\dfrac{5}{\pi}~\text{Hz}$
- $\dfrac{10}{\pi}~\text{Hz}$
- $\dfrac{\pi}{5}~\text{Hz}$
- $\pi~\text{Hz}$
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The Correct Option is B
Solution and Explanation
Total mass after collision $= 1 + 0.5 = 1.5~\text{kg}$
Frequency $f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}} = \dfrac{1}{2\pi} \sqrt{\dfrac{600}{1.5}}$
$= \dfrac{1}{2\pi} \cdot \sqrt{400} = \dfrac{20}{2\pi} = \dfrac{10}{\pi}~\text{Hz}$
Frequency $f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}} = \dfrac{1}{2\pi} \sqrt{\dfrac{600}{1.5}}$
$= \dfrac{1}{2\pi} \cdot \sqrt{400} = \dfrac{20}{2\pi} = \dfrac{10}{\pi}~\text{Hz}$
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