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Standard XII

Physics

U Tube SHM

A mass M is s...

Question

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 simple harmonic oscillations with a time period T. If the mass is increased by m then the time period becomes $(\frac{5}{4} T)$ . The ratio of $\frac{m}{M}$ is

9/16

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25/16

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4/5

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5/4

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Solution

The correct option is A
9/16

T = $2 π \sqrt{\frac{m}{k}} \Rightarrow m \propto T^{2} \Rightarrow \frac{m_{2}}{m_{1}} = \frac{T_{2}^{2}}{T_{1}^{2}}$

$\Rightarrow \frac{M + m}{M} = {(\frac{\frac{5}{4} T}{T})}^{2} \Rightarrow \frac{m}{M} = \frac{9}{16}$

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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 simple harmonic oscillations with a time period T. If the mass is increased by m then the time period becomes (54T) . The ratio of mM is

The correct option is A 9/16 T = 2π√mk⇒m∝T2⇒m2m1=T22T21 ⇒M+mM=(54TT)2⇒mM=916

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 simple harmonic oscillations with a time period T. If the mass is increased by m then the time period becomes $(\frac{5}{4} T)$ . The ratio of $\frac{m}{M}$ is

The correct option is A
9/16

T = $2 π \sqrt{\frac{m}{k}} \Rightarrow m \propto T^{2} \Rightarrow \frac{m_{2}}{m_{1}} = \frac{T_{2}^{2}}{T_{1}^{2}}$

$\Rightarrow \frac{M + m}{M} = {(\frac{\frac{5}{4} T}{T})}^{2} \Rightarrow \frac{m}{M} = \frac{9}{16}$