Question

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

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

Answer 1

The correct option is C $0.37$

Given:

Frequency of oscillation reduced by $20\%$ on adding m

To find:

Ratio of $\frac{m}{M}=?$

Sol:

the frequency of oscillation is given by

$f=\frac{K}{\sqrt{I}}$

$f_1=\frac{K}{\sqrt{\frac{M(2L{)}^2}{12}}}$

$f_2=\frac{K}{\sqrt{\frac{M(2L{)}^2}{12}+2m{\left(\frac{L}{2}\right)}^2}}$

$f_2=0.8f_1$

$\frac{K}{\sqrt{\frac{M(2L{)}^2}{12}+2m{\left(\frac{L}{2}\right)}^2}}=\frac{0.8}{10}\times \frac{K}{\sqrt{\frac{M(2L{)}^2}{12}}}$

$16[\frac{M(2L{)}^2}{12}+2m{\left(\frac{L}{2}\right)}^2]=2.5.\frac{M(2L{)}^2}{12}$.

$16\times 2m{\left(\frac{L}{2}\right)}^2=9\frac{M(2L{)}^2}{12}$

$\frac{m}{M}=\frac{9\times 4L^2\times 4}{12\times 16\times 2\times L^2}$

$\frac{m}{M}=0.375$