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 mass m, are attached at…

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 mass m, are attached at a 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

MARKS Content Team · Accepted Answer

Correct answer: (D) 0.37. Let C be the torsional constant of the wire. f=1/(2π)√(CM2L^212)=1/(2π)√(3CM. L^2) After masses are attached, f^(′)=1/(2π) √(CM.2L^212+mL^24×2) ⇒0.8f=1/(2π) √(CM3+m2L^2) ⇒0.64×3C/M=C/M3+m/2 0.64M+0.64×3/2m=M ⇒m/M=0.37

A rod of mass $M$ and length $2 L$ is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses, each of mass $m$ , are attached at a 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

Solution

More Oscillations Questions