Question

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 a distance ${L \mathord{\left/
 {\vphantom {L 2}} \right.} 2}$ from its center on both sides, it reduces the oscillation frequency by $20\% $. The value of ${m \mathord{\left/
 {\vphantom {m M}} \right.} M}$ is close to:
(A) $0.175$
(B) $0.375$
(C) $0.575$
(D) $0.775$

Answer 1

Hint
From the formula for frequency for torsional oscillations, we find the frequency without the masses and then with the masses. As the frequency in the second case is $20\% $ less than the first case, we can equate them and find the answer.
In this solution, we will be using the following formula,
$\Rightarrow f = \dfrac{k}{{\sqrt I }}$
where $f$ is the frequency of oscillation and $I$ is the moment of inertia of a body.
and $k$ is the torsional constant.

Complete step by step answer
In the question we are given that a rod of length $2L$ and mass $M$ is suspended by a wire. Therefore the moment of inertia of the rod about the wire will be
$\Rightarrow I = \dfrac{{M{{\left( {2L} \right)}^2}}}{{12}}$
This gives us,
$\Rightarrow I = \dfrac{{4M{L^2}}}{{12}}$
Cancelling 4 from numerator and denominator,
$\Rightarrow I = \dfrac{{M{L^2}}}{3}$
So the frequency of oscillation in the first case will be given by,
$\Rightarrow {f_1} = \dfrac{k}{{\sqrt I }}$
Substituting the value of $I$ we get,
$\Rightarrow {f_1} = \dfrac{k}{{\sqrt {\dfrac{{M{L^2}}}{3}} }}$
Now for the second case, 2 masses $m$ were attached on either side of the rod at a distance of ${L \mathord{\left/
 {\vphantom {L 2}} \right.
} 2}$ from the center. Hence, the new moment of inertia will be,
$\Rightarrow I' = \dfrac{{M{L^2}}}{3} + m{\left( {\dfrac{L}{2}} \right)^2} + m{\left( {\dfrac{L}{2}} \right)^2}$
So this can be simplified as,
$\Rightarrow I' = \dfrac{{M{L^2}}}{3} + 2m\dfrac{{{L^2}}}{4}$
On cancelling the 2 from numerator and denominator, we get
$\Rightarrow I' = \dfrac{{M{L^2}}}{3} + \dfrac{{m{L^2}}}{2}$
So in this case the frequency will be,
$\Rightarrow {f_2} = \dfrac{k}{{\sqrt {I'} }}$
Substituting we get,
$\Rightarrow {f_2} = \dfrac{k}{{\sqrt {\dfrac{{M{L^2}}}{3} + \dfrac{{m{L^2}}}{2}} }}$
Now according to the problem the frequency in the second case gets reduced $20\% $. So we can write,
$\Rightarrow {f_2} = \dfrac{{80}}{{100}}{f_1}$
This gives us the frequency as,
$\Rightarrow {f_2} = \dfrac{8}{{10}}{f_1}$
Now substituting the values we get
$\Rightarrow \dfrac{k}{{\sqrt {\dfrac{{M{L^2}}}{3} + \dfrac{{m{L^2}}}{2}} }} = \dfrac{8}{{10}}\dfrac{k}{{\sqrt {\dfrac{{M{L^2}}}{3}} }}$
We can cancel the $k$ from both sides and then on doing cross multiplication, we get
$\Rightarrow \sqrt {\dfrac{{M{L^2}}}{3}} = \dfrac{8}{{10}}\sqrt {\dfrac{{M{L^2}}}{3} + \dfrac{{m{L^2}}}{2}} $
Now squaring on both the sides we have,
$\Rightarrow \dfrac{{M{L^2}}}{3} = \dfrac{{64}}{{100}}\left( {\dfrac{{M{L^2}}}{3} + \dfrac{{m{L^2}}}{2}} \right)$
On opening the bracket we get,
$\Rightarrow \dfrac{{M{L^2}}}{3} = \dfrac{{64}}{{100}} \times \dfrac{{M{L^2}}}{3} + \dfrac{{64}}{{100}} \times \dfrac{{m{L^2}}}{2}$
On simplifying and taking the common terms on one side we get
$\Rightarrow \dfrac{{M{L^2}}}{3} - \dfrac{{64M{L^2}}}{{300}} = \dfrac{{32m{L^2}}}{{100}}$
On doing the calculation in the LHS
$\Rightarrow \dfrac{{\left( {100 - 64} \right)M{L^2}}}{{300}} = \dfrac{{32m{L^2}}}{{100}}$
We can cancel the $\dfrac{{{L^2}}}{{100}}$ from both sides and get
$\Rightarrow \dfrac{{36M}}{3} = 32m$
Now we bring $M$ to the RHS and taking the rest to the LHS,
$\Rightarrow \dfrac{m}{M} = \dfrac{{36}}{{3 \times 32}}$
On doing the calculation we get a value of $\dfrac{m}{M} = 0.375$
So the correct answer is option (B).

Note
The moment of inertia of a body is its tendency to resist angular acceleration. The moment of inertia of any body can be calculated by the sum of product of the masses of the particles in the body and their perpendicular distance from the axis.