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:

We have a uniform rod of mass $$M$$ and length $$2L$$ hung at its mid-point by a thin wire, so it performs torsional oscillations about a vertical axis through its centre. The oscillation frequency of a torsional pendulum is determined by the formula

$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{I}},$$

where $$k$$ is the torsional constant of the wire and $$I$$ is the moment of inertia of the system about the axis of rotation.

For the bare rod the moment of inertia about a perpendicular axis through its centre is obtained from the standard result for a slender rod:

$$I_0=\int x^2\,dm=\dfrac{1}{12}M(2L)^2=\dfrac{1}{3}ML^{2}.$$

Hence the original frequency is

$$f_0=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{I_0}}.$$

Now two small masses, each of mass $$m$$, are attached symmetrically at a distance $$\dfrac{L}{2}$$ from the centre. The contribution of each attached mass to the moment of inertia is $$mr^{2}=m\left(\dfrac{L}{2}\right)^{2}=\dfrac{mL^{2}}{4}$$, so for both masses together we add

$$I_{\text{masses}}=2\times\dfrac{mL^{2}}{4}=\dfrac{mL^{2}}{2}.$$

The new total moment of inertia therefore becomes

$$I'=I_0+\dfrac{mL^{2}}{2}=\dfrac{1}{3}ML^{2}+\dfrac{1}{2}mL^{2}.$$

Because the wire is unchanged, the torsional constant $$k$$ remains the same. The modified frequency is

$$f'=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{I'}}.$$

According to the statement of the problem, attaching the masses reduces the frequency by 20 %, i.e.

$$f' = 0.8\,f_0.$$

Taking the ratio of the expressions for $$f'$$ and $$f_0$$ we get

$$\dfrac{f'}{f_0}=\sqrt{\dfrac{I_0}{I'}}=0.8.$$

Squaring both sides gives

$$\dfrac{I_0}{I'}=0.64 \quad\Longrightarrow\quad I'=\dfrac{I_0}{0.64}=\dfrac{25}{16}I_0.$$

Substituting $$I'$$ and $$I_0$$ explicitly,

$$\dfrac{1}{3}ML^{2}+\dfrac{1}{2}mL^{2}=\dfrac{25}{16}\left(\dfrac{1}{3}ML^{2}\right).$$

Cancelling the common factor $$L^{2}$$ we have

$$\dfrac{1}{3}M+\dfrac{1}{2}m=\dfrac{25}{48}M.$$

Rearranging to isolate $$m$$,

$$\dfrac{1}{2}m=\dfrac{25}{48}M-\dfrac{1}{3}M =\left(\dfrac{25}{48}-\dfrac{16}{48}\right)M =\dfrac{9}{48}M =\dfrac{3}{16}M.$$

Multiplying by 2,

$$m=\dfrac{3}{8}M.$$

Therefore the sought ratio is

$$\dfrac{m}{M}=\dfrac{3}{8}=0.375\approx 0.37.$$

Hence, the correct answer is Option D.