Question

A body executes simple harmonic motion under the action of a force $F_1$ with a time period $\left(4/5\right)$ seconds. If the force is changed to $F_2$, it executes SHM with time period $\left(3/5\right)$ seconds. 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. $12/25$
• B. $24/25$
• C. $35/24$
• D. $25/12$

Answer 1

The correct option is A $12/25$

$F_1=-k_{1}x,F_2=-k_{2}x$
$F_1=-\left(\frac{k_1}{m}\right)x;F_2=-\left(\frac{k_2}{m}\right)x$
$\Rightarrow F_1=-{\omega }_1^{2}x;f_2=-{\omega }_2^{2}x$
Now, resultant force $F=F_1+F_2=-k_{1}x-k_{2}x$
$\Rightarrow -kx=-k_{1}x-k_{2}x$
$\Rightarrow k=k_1+k_2\Rightarrow m{\omega }^2=m{\omega }_1^2+m{\omega }_2^2$
$\Rightarrow {\omega }^2={\omega }_1^2+{\omega }_2^2$
$\Rightarrow {\left(\frac{2\pi }{T}\right)}^2={\left(\frac{2\pi }{T_1}\right)}^2+{\left(\frac{2\pi }{T_2}\right)}^2$
$\Rightarrow \frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}$
$\therefore T=\frac{T_1{T}_2}{\sqrt{T_1^2+T_2^2}}$
$=\frac{\frac{4}{5}\times \frac{3}{5}}{\sqrt{{\left(\frac{4}{5}\right)}^2+{\left(\frac{3}{5}\right)}^2}}=\frac{12}{25}\text{second}$
Hence, the correct answer is option (a).