(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 $\left(\frac{M}{m}\right)$?

MHT CET 2020 All MHT CET

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 $\left(\frac{M}{m}\right)$?

Similar Questions

$A$ force of $6.4 \, N$ stretches a vertical spring by $0.1 \, m$. The mass that must be suspended from the spring so that it oscillates with a period of $\left( \frac{\pi}{4} \right) \, s$ is ... $kg$.

Two identical point masses $P$ and $Q$,suspended from two separate massless springs of spring constants $k_1$ and $k_2$,respectively,oscillate vertically. If their maximum speeds are the same,the ratio $(A_Q / A_P)$ of the amplitude $A_Q$ of mass $Q$ to the amplitude $A_P$ of mass $P$ is

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