A horizontal spring executes S.H.M. with ampl | vedclass

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Question

(A) When mass $m_{1}$ is attached to the spring,the total energy of the $S.H.M.$ is $E = \frac{1}{2} k A_{1}^{2}$.At the mean position,the velocity $v

Vedclass · Accepted Answer

$\left[\frac{m_{1}}{m_{1}+m_{2}}\right]^{1 / 2}$

Vedclass · Answer

(A) When mass $m_{1}$ is attached to the spring,the total energy of the $S.H.M.$ is $E = \frac{1}{2} k A_{1}^{2}$.At the mean position,the velocity $v_{1}$ of the mass $m_{1}$ is maximum,given by $v_{1} = \omega_{1} A_{1} = \sqrt{\frac{k}{m_{1}}} A_{1}$.When mass $m_{2}$ is placed on $m_{1}$ at the mean position,the momentum of the system is conserved because there is no external horizontal force.Initial momentum $p_{i} = m_{1} v_{1}$.Final momentum $p_{f} = (m_{1} + m_{2}) v_{2}$,where $v_{2}$ is the new velocity at the mean position.Since $p_{i} = p_{f}$,we have $m_{1} v_{1} = (m_{1} + m_{2}) v_{2}$.$v_{2} = \frac{m_{1}}{m_{1} + m_{2}} v_{1}$.The new angular frequency is $\omega_{2} = \sqrt{\frac{k}{m_{1} + m_{2}}}$.Since $v_{2} = \omega_{2} A_{2}$,we have $A_{2} = \frac{v_{2}}{\omega_{2}} = \frac{m_{1} v_{1}}{(m_{1} + m_{2})} \sqrt{\frac{m_{1} + m_{2}}{k}}$.Substituting $v_{1} = \sqrt{\frac{k}{m_{1}}} A_{1}$,we get $A_{2} = \frac{m_{1}}{(m_{1} + m_{2})} \sqrt{\frac{k}{m_{1}}} A_{1} \sqrt{\frac{m_{1} + m_{2}}{k}} = A_{1} \sqrt{\frac{m_{1}}{m_{1} + m_{2}}}$.Therefore,$\frac{A_{2}}{A_{1}} = \left[\frac{m_{1}}{m_{1} + m_{2}}\right]^{1/2}$.

$A$ horizontal spring executes $S.H.M.$ with amplitude $A_{1}$,when mass $m_{1}$ is attached to it. When it passes through the mean position,another mass $m_{2}$ is placed on it. Both masses move together with amplitude $A_{2}$. Therefore,the ratio $A_{2}: A_{1}$ is:

Similar Questions

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