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JEE Main 2019 (9 Jan, Shift 2) · Other SHM Systems

Other SHM Systems question from JEE Main 2019 (9 Jan, Shift 2)

Q16JEE Main 2019 · 9 JAN · SHIFT 2Difficult

Other SHM Systems

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 attached at 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:

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Answer: (c)

  1. Torsional pendulum: f∝1If \propto \frac{1}{\sqrt{I}}. I0=M(2L)212=ML23I_0 = \frac{M(2L)^2}{12} = \frac{ML^2}{3}.
  2. With the masses: I1=ML23+2m(L2)2=ML23+mL22I_1 = \frac{ML^2}{3} + 2m\left(\frac{L}{2}\right)^2 = \frac{ML^2}{3} + \frac{mL^2}{2}.
  3. f1=0.8f0⇒I1I0=10.64=1.5625f_1 = 0.8f_0 \Rightarrow \frac{I_1}{I_0} = \frac{1}{0.64} = 1.5625
  4. 3m2M=0.5625⇒mM=0.375≈0.37\frac{3m}{2M} = 0.5625 \Rightarrow \frac{m}{M} = 0.375 \approx 0.37, answer (c).
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