Oscillations

JEE Main level, single correct, medium

190 of 194

Question 190 of 194

A rod of mass MM and length 2L2L is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses, each of mass mm, are attached at a distance L/2L/2 from its centre on both sides, it reduces the oscillation frequency by 2020%. The value of ratio m/Mm/M is close to

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Answer: (D) 0.370.37

Let CC be the torsional constant of the wire.

f=12πCM(2L)212=12π3CM.L2f=\dfrac{1}{2\pi }\sqrt{\dfrac{C}{\dfrac{M{\left(2L\right)}^{2}}{12}}}=\dfrac{1}{2\pi }\sqrt{\dfrac{3C}{M.{L}^{2}}}

After masses are attached,

f′=12πCM.(2L)212+mL24×2{f}^{'}=\dfrac{1}{2\pi }\sqrt{\dfrac{C}{\dfrac{M.{\left(2L\right)}^{2}}{12}+\dfrac{m{L}^{2}}{4}\times 2}}

⇒0.8f=12πC(M3+m2)L2\Rightarrow 0.8f=\dfrac{1}{2\pi }\sqrt{\dfrac{C}{\left(\dfrac{M}{3}+\dfrac{m}{2}\right){L}^{2}}}

⇒0.64×3CM=CM3+m2\Rightarrow 0.64\times \dfrac{3C}{M}=\dfrac{C}{\dfrac{M}{3}+\dfrac{m}{2}}

(0.64)M+0.64×32m=M\left(0.64\right)M+0.64\times \dfrac{3}{2}m=M

⇒mM=0.37\Rightarrow \dfrac{m}{M}=0.37

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