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Standard XII

Physics

Forced Oscillations

A rod of mass...

Question

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 oscillation frequency by $20 %$ . The value of ratio m/M is close to :

0.175

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0.375

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0.575

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0.775

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Solution

The correct option is B 0.375
Frequency of torsonal oscillations is given by
$f = \frac{k}{\sqrt{I}}$
$f_{1} = \frac{k}{\sqrt{\frac{M (2 L)^{2}}{12}}}$
$f_{2} = \frac{k}{\sqrt{\frac{M (2 L)^{2}}{12} + 2 m {(\frac{L}{2})}^{2}}}$
$f_{2} = 0.8 f_{1}$
$\frac{m}{M} = 0.375$

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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' areattached at distance ′L/2′ from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to :

The correct option is B 0.375Frequency of torsonal oscillations is given by f=k√If1=k√M(2L)212f2=k√M(2L)212+2m(L2)2f2=0.8f1mM=0.375

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 oscillation frequency by $20 %$ . The value of ratio m/M is close to :

The correct option is B 0.375
Frequency of torsonal oscillations is given by
$f = \frac{k}{\sqrt{I}}$
$f_{1} = \frac{k}{\sqrt{\frac{M (2 L)^{2}}{12}}}$
$f_{2} = \frac{k}{\sqrt{\frac{M (2 L)^{2}}{12} + 2 m {(\frac{L}{2})}^{2}}}$
$f_{2} = 0.8 f_{1}$
$\frac{m}{M} = 0.375$