SATHEE: Simple Harmonic Motion 4 Question 17

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. $ω = \sqrt{\frac{k}{m_{2}}}, A = \frac{m_{1} g}{k}$

Solution:

When $m_{1}$ is removed only $m_{2}$ is left. Therefore, angular frequency $ω = \sqrt{k / m_{2}}$

Let $x_{1}$ be the extension when only $m_{2}$ is left. Then,

$k x_{1} = m_{2} g or x_{1} = \frac{m_{2} g}{k}$

Similarly, let $x_{2}$ be the extension in equilibrium when both $m_{1}$ and $m_{2}$ are suspended. Then,

$\begin{aligned} (m_{1} + m_{2}) g & = k x_{2} \\ \Rightarrow x_{2} & = \frac{(m_{1} + m_{2}) g}{k} \end{aligned}$

From Eqs. (i) and (ii), amplitude of oscillation

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Simple Harmonic Motion 4 Question 17

17. Two masses m1 and m2 are suspended together by a massless spring of spring constant k (Fig.). When the masses are in equilibrium, m1 is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of m2.

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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 $m_{2}$ .