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PhysicsSHMCalculation of time periodEasy2 minQB

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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 SHM of time period $T$ . If the mass is increased by $m$ , the time period become $\frac{5 T}{3}$ . Then the ratio of $\frac{m}{M}$ is

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Answer:

Solution:

$T = 2 π \sqrt{\frac{M}{k}}$ when mass is increased by $m$ then ...(i)

$T = 2 π \sqrt{\frac{M + m}{k}}$

$\Rightarrow$ $\frac{5 T}{3} = 2 π \sqrt{\frac{M + m}{k}}$ ...(ii)

Dividing Eq. (i) by Eq. (ii), we get

$\frac{3}{5} = \sqrt{\frac{M}{M + m}}$

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

$\Rightarrow$ $9 M + 9 m = 25 M$

$\Rightarrow$ $16 M = 9 m$

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

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: QB

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