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PhysicsSHMAngular SHMMedium2 minPYQ_2019
PhysicsMediumsingle choice

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 massm, are attached at a distance L/2 from its centre on both sides, it reduces the oscillation frequency by20%. The value of ratiom/Mis close to

Options:

Answer:
D
Solution:

Let C be the torsional constant of the wire.

f=12πCM2L212=12π3CM. L2

After masses are attached,

f=12π CM.2L212+mL24×2

0.8f=12π CM3+m2L2

0.64×3CM=CM3+m2

0.64M+0.64×32m=M

mM=0.37

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Angular SHM
2mℹ️ Source: PYQ_2019

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