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  • Help me please, A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillations frequency by 20%. The value of ratio m/M is close to:

 

  • Option 1)

    0.57

  • Option 2)

     

    0.17

  • Option 3)

    0.37

  • Option 4)

    0.77

Answers (1)

best_answer

 

Time Period of Torsional pendulum case -

T=2\pi \sqrt{\frac{I}{K}}

- wherein

I= moment of inertia

K= torsional constant

frequency  f=\frac{K}{\sqrt I}

f_{1}=\frac{K}{\sqrt{(\frac{M(2L)^{2}}{12})}}

f_{2}=\frac{K}{\sqrt{(\frac{M(2L)^{2}}{12})+2m(\frac{L}{2})^{2}}}

f_{2}=0.8f_{1}

\frac{m}{M}=0.375

 

 


Option 1)

0.57

Option 2)

 

0.17

Option 3)

0.37

Option 4)

0.77

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