Question

A body executes simple harmonic motion under the action of force $F_{1}$ with a time period $\frac{4}{5} s$. If the force is changed to $F_{2}$ it executes simple harmonic motion with time period $\frac{3}{5} s$. If both forces $F_{1}$ and $F_{2}$ act simultaneously in the same direction on the body, its time period will be

• A. $ \frac{12}{25}\,s $
• B. $ \frac{24}{25}\,s $
• C. $ \frac{35}{24}\,s $
• D. $ \frac{15}{12}\,s $

Answer 1

Answer: (A)

Under the influence of one force $F_{1}=m \omega_{1}^{2} y$
and under the action of another force, $F_{2}=m \omega_{2}^{2} y$
Under the action of both the forces $F=F_{1}+F_{2}$
$\Rightarrow m \omega^{2} y=m \omega_{1}^{2} y+m \omega_{2}^{2} y$
$\Rightarrow \omega^{2} =\omega_{1}^{2}+\omega_{2}^{2} $
$ \Rightarrow \left[\frac{2 \pi}{T}\right]^{2} =\left[\frac{2 \pi}{T_{1}}\right]^{2}+\left[\frac{2 \pi}{T_{2}}\right]^{2} $
$\Rightarrow T=\sqrt{\frac{T_{1}^{2} \times T_{2}^{2}}{T_{1}^{2}+T_{2}^{2}}}$
$=\sqrt{\frac{\left(\frac{4}{5}\right)^{2}\left(\frac{3}{5}\right)^{2}}{\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}}}$
$=\frac{12}{25} s$