Let C be the torsional constant of the wire.
f=2π112M(2L)2C=2π1M.L23C
After masses are attached,
f′=2π112M.(2L)2+4mL2×2C
⇒0.8f=2π1(3M+2m)L2C
⇒0.64×M3C=3M+2mC
(0.64)M+0.64×23m=M
⇒Mm=0.37
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…
JEE Main 2019 — Physics Waves & Oscillations
2019mcqmedium
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
Official previous-year question
Held on 9 Jan 2019 · Verified 6 Jul 2026.
Options
- A
0.17
- B
0.77
- C
0.57
- D
0.37
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