Question

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

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

Answer 1

Answer: (A)

$F_{1}=-k_{1}x;F_{2}=-k_{2}x$
$a_{1}=-\left(\frac{k_{1}}{m}\right)x;a_{2}=-\left(\frac{k_{2}}{m}\right)x$
Also, $a_{1}=-\omega _{1}^{2}x;a_{2}=-\omega _{2}^{2}x$
Now, resultant force $F=F_{1}+F_{2}=-k_{1}x-k_{2}x$
$-kx=-k_{1}x-k_{2}x$
$k=k_{1}+k_{2}$
$m\omega ^{2}=m\omega _{1}^{2}+m\omega _{2}^{2}$
$\omega ^{2}=\omega _{1}^{2}+\omega _{2}^{2}$
$\left(\frac{2 \pi }{T}\right)^{2}=\left(\frac{2 \pi }{T_{1}}\right)^{2}+\left(\frac{2 \pi }{T_{2}}\right)^{2}$
$\frac{1}{T^{2}}=\frac{1}{T_{1}^{2}}+\frac{1}{T_{2}^{2}}$
$\therefore T=\frac{T_{1} T_{2}}{\sqrt{T_{1}^{2} + T_{2}^{2}}}$
$\Rightarrow T=\frac{\frac{4}{5} \times \frac{3}{5}}{\sqrt{\left(\frac{4}{5}\right)^{2} + \left(\frac{3}{5}\right)^{2}}}=\frac{12}{25}s$