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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’ 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
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	0.17plz eaplain....thnx


8 months ago

Arun
25768 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

8 months ago
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