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Grade: 12th pass

                        

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 at a distance L/2 from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio of m/M is close to 0.77 0.57 0.37 0.17 plz eaplain.... thnx

5 months ago

Answers : (1)

Arun
24742 Points
							
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
 
 
5 months ago
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