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 $F=F_1+F_2=-k_{1}x-k_{2}x$
$-kx=-k_{1}x-k_{2}x$
$k=k_1+k_2$
$m{\omega }^2=m{\omega }_1^2+m{\omega }_2^2$
${\omega }^2={\omega }_1^2+{\omega }_2^2$
${\left(\frac{2\pi }{T}\right)}^2={\left(\frac{2\pi }{T_1}\right)}^2+{\left(\frac{2\pi }{T_2}\right)}^2$
$\frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}$
$\therefore T=\frac{T_1{T}_2}{\sqrt{T_1^2+T_2^2}}$
$\Rightarrow T=\frac{\frac{4}{5}\times \frac{3}{5}}{\sqrt{{\left(\frac{4}{5}\right)}^2+{\left(\frac{3}{5}\right)}^2}}=\frac{12}{25}s$