Question

A rod of mass $M$ and length $2L$ is supended at its middle by a wire. It exhibits torsional oscillations. If two masses each of $m$ are attached at distance $\frac{L}{2}$ from its centre on both sides, it reduces the oscillation frequency by $20\%$. The value of ratio $\frac{m}{M}$ is close to

• A. $0.77$
• B. $0.17$
• C. $0.57$
• D. $0.37$

Answer 1

The correct option is D $0.37$


Frequency of oscillating is given by-
$f=\frac{1}{2\pi }\sqrt{\frac{C}{I}}$

Let $f_1$ is initial oscillating frequency and $f_2$ is the final oscillating frequency after placing masses on rod.

$f_1=\frac{1}{2\pi }\sqrt{\frac{C}{I}}$

Moment of inertia for rod mass $M$
$\Rightarrow I=\frac{m{r}^2}{12}$
$\Rightarrow I=\frac{m(2L{)}^2}{12}$
$\Rightarrow I=\frac{m{L}^2}{3}$

So,
$f_1=\frac{1}{2\pi }\sqrt{\frac{3C}{M{L}^2}}.....\left(1\right)$

After placing two $m$ mass block on rod
$f_2=\frac{1}{2\pi }\sqrt{\frac{C}{I^{\prime }}}$

New moment of inertia
$I^{\prime }=\frac{M{L}^2}{3}+m(L/2{)}^2+m(L/2{)}^2$
$I^{\prime }=\frac{M{L}^2}{3}+\frac{m{L}^2}{2}$

So,
$f_2=\frac{1}{2\pi }\sqrt{\frac{C}{\frac{M{L}^2}{3}+\frac{m{L}^2}{2}}}.....\left(2\right)$

Given: frequency reduces by $80\%$
$\frac{(f_1-f_2)}{f_1}\times 100=20$
$\frac{f_2}{f_1}=0.8.....\left(3\right)$

Substitute the values of $f_1$ and $f_2$ in eq.3

$\frac{\frac{1}{2\pi }\sqrt{\frac{C}{\frac{M{L}^2}{3}+\frac{m{L}^2}{2}}}}{\frac{1}{2\pi }\sqrt{\frac{3C}{M{L}^2}}}=0.8$

$\frac{M}{3(\frac{M}{3}+\frac{m}{2})}=0.64$

$M=0.64M+0.96m$

$\frac{m}{M}=\frac{0.36}{0.96}=0.37$