MHT CET · Physics · Oscillations

A mass $m_{1}$ connected to a horizontal spring performs S.H.M. with amplitude $A$ . While mass $m_{1}$ is passing through mean position another mass $m_{2}$ is placed on it so that both the masses move together with amplitude $A_{1}$ . The ratio of $\frac{A_{1}}{A}$ is $(m_{2} < m_{1})$

A ${[\frac{m_{1}}{m_{1} + m_{2}}]}^{\frac{1}{2}}$
B ${[\frac{m_{1} + m_{2}}{m_{1}}]}^{\frac{1}{2}}$
C ${[\frac{m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}}$
D ${[\frac{m_{1} + m_{2}}{m_{2}}]}^{\frac{1}{2}}$

Verified Solution

Answer & Solution

Correct Answer

(A) ${[\frac{m_{1}}{m_{1} + m_{2}}]}^{\frac{1}{2}}$

Step-by-step Solution

Detailed explanation

P.E of the oscillating mass is given by

E = \frac{1}{2} m ω x^{2} = \frac{1}{2} k x^{2}

where

k = m ω^{2}

Let

k = m ω^{2}

be the velocity at mean position. When mass

M_{2}

is attached

v_{2} = \frac{m_{1} v_{1}}{m_{1} + m_{2}}

A_{1}

be the new amplitude

\frac{1}{2} k A_{1}^{2} = \frac{1}{2} (m_{1} + m_{2}) v_{2}^{2}

\frac{1}{2} (m_{1} + m_{2}) \frac{m_{1}^{2} v_{1}^{2}}{{(m_{1} + m_{2})}^{2}}

= \frac{1}{2} m_{1} v_{1}^{2} \times \frac{m_{1}}{m_{1} + m_{2}}

= \frac{1}{2} k A^{2} \times \frac{m_{1}}{m_{1} + m_{2}}

\Rightarrow \frac{A_{1}}{A} = \sqrt{\frac{m_{1}}{m_{1} + m_{2}}}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A mass m1 connected to a horizontal spring performs S.H.M. with amplitude A. While mass m1 is passing through mean position another mass m2 is placed on it so that both the masses move together with amplitude A1. The ratio of A1A is (m2<m1)

Step-by-step Solution

A mass $m_{1}$ connected to a horizontal spring performs S.H.M. with amplitude $A$ . While mass $m_{1}$ is passing through mean position another mass $m_{2}$ is placed on it so that both the masses move together with amplitude $A_{1}$ . The ratio of $\frac{A_{1}}{A}$ is $(m_{2} < m_{1})$