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)

$\frac{12}{25}\,s$

Answer 2

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 $F=F_{1}+F_{2}$ $\Rightarrow m \omega^{2} y=m \omega_{1}^{2} y+m \omega_{2}^{2} y$ $\Rightarrow \omega^{2} =\omega_{1}^{2}+\omega_{2}^{2}$ $\Rightarrow \left[\frac{2 \pi}{T}\right]^{2} =\left[\frac{2 \pi}{T_{1}}\right]^{2}+\left[\frac{2 \pi}{T_{2}}\right]^{2}$ $\Rightarrow T=\sqrt{\frac{T_{1}^{2} \times T_{2}^{2}}{T_{1}^{2}+T_{2}^{2}}}$ $=\sqrt{\frac{\left(\frac{4}{5}\right)^{2}\left(\frac{3}{5}\right)^{2}}{\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}}}$ $=\frac{12}{25} s$