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Physics - SHM Question with Solution | TestHub

PhysicsSHMMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

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 SHM of time period $T$ . If the mass is increased by $m$ , the time period becomes $5 T / 3$ . Then the ratio of $m / M$ is

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Answer:

Solution:

Initially, $T = 2 π \sqrt{M / k}$

Finally, $\frac{5 T}{3} = 2 π \sqrt{\frac{M + m}{k}}$

$∴$ $\frac{5}{3} \times 2 π \sqrt{\frac{M}{k}} = 2 π \sqrt{\frac{M + m}{k}}$

or $\frac{25}{9} \frac{M}{k} = \frac{M + m}{k}$

or $9 m + 9 M = 25M$

or $\frac{m}{M} = \frac{16}{9}$ .

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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