(D) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{M}{k}}$.When the mass is increased by $m$,the new time period $T'$ is gi

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(D) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{M}{k}}$.When the mass is increased by $m$,the new time period $T'$ is given as $T' = \frac{5T}{3}$.Thus,$\frac{5T}{3} = 2\pi \sqrt{\frac{M+m}{k}}$.Substituting $T = 2\pi \sqrt{\frac{M}{k}}$ into the equation:$\frac{5}{3} \times 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M+m}{k}}$.Squaring both sides:$\frac{25}{9} \times \frac{M}{k} = \frac{M+m}{k}$.$\frac{25}{9}M = M + m$.Dividing by $M$:$\frac{25}{9} = 1 + \frac{m}{M}$.$\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}$.Therefore,the ratio $\frac{M}{m} = \frac{9}{16}$.

Class 11 Physics Oscillations SHM of Spring Mass System

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$A$ mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of period $T$. If the mass is increased by $m$,the time period becomes $\frac{5T}{3}$. What is the ratio $\left(\frac{M}{m}\right)$?

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A
$\frac{25}{9}$
B
$\frac{16}{9}$
C
$\frac{9}{25}$
D
$\frac{9}{16}$

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$A$ mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of period $T$. If the mass is increased by $m$,the time period becomes $\frac{5T}{3}$. What is the ratio $\left(\frac{M}{m}\right)$?

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$A$ $2\, kg$ block moving with $10\, m/s$ strikes a spring of constant $\pi^2\, N/m$ attached to a $2\, kg$ block at rest kept on a smooth floor. The time for which the moving block remains in contact with the spring will be ... $\sec$.

$A$ mass $m$ performs oscillations of period $T$ when suspended by a spring of force constant $K$. If the spring is cut into two equal parts and arranged in parallel,and the same mass $m$ is oscillated by them,then the new time period will be:

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