Simple Harmonic Motion 4 Question 17
17. Two masses $m _1$ and $m _2$ are suspended together by a massless spring of spring constant $k$ (Fig.). When the masses are in equilibrium, $m _1$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of $m _2$.
$(1981,3 M)$
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Answer:
Correct Answer: 17. $\omega=\sqrt{\frac{k}{m _2}}, A=\frac{m _1 g}{k}$
Solution:
- When $m _1$ is removed only $m _2$ is left. Therefore, angular frequency $\omega=\sqrt{k / m _2}$
Let $x _1$ be the extension when only $m _2$ is left. Then,
$$ k x _1=m _2 g \text { or } x _1=\frac{m _2 g}{k} $$
Similarly, let $x _2$ be the extension in equilibrium when both $m _1$ and $m _2$ are suspended. Then,
$$ \begin{aligned} \left(m _1+m _2\right) g & =k x _2 \\ \Rightarrow \quad x _2 & =\frac{\left(m _1+m _2\right) g}{k} \end{aligned} $$
From Eqs. (i) and (ii), amplitude of oscillation
