Simple Harmonic Motion Ques 40
- 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: 40.$(\omega=\sqrt{\frac{k}{m _2}}, A=\frac{m _1 g}{k})$
Solution:
Formula:
- 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} $ $\quad$ …….(i)
Similarly, let $x _2$ be the extension in equilibrium when both $m _1$ and $m _2$ are suspended. Then,
$\left(m _1+m _2\right) g =k x _2 $
$\Rightarrow \quad x _2 =\frac{\left(m _1+m _2\right) g}{k}$ $\quad$ …….(ii)
From Eqs. (i) and (ii), amplitude of oscillation
$A= x_2 -x_1= \frac{m_1 g}{k}$
BETA
