Question

A body executes S.H.M. under the action of a force $F$${}_1$ with a time period $4/5$ seconds. If the force is changed to $F$${}_2$, it executes S.H.M. with a time period $3/5$ seconds. If both the forces $F$${}_1$and $F$${}_2$ act simultaneously in the same direction on the body, then its time period in seconds is:

• A. $12/25$
• B. $24/25$
• C. $25/24$
• D. $25/12$

Answer 1

The correct option is A $12/25$

$F_1=m{\omega }^{2}x$
$T=\frac{2\pi }{\omega }$
$F=ma=$ $4m{\pi }^{2}x\frac{1}{T^2}$
$\therefore F_1=\frac{4m{\pi }^2}{T_1^2}x$
$F_2=\frac{4m{\pi }^2}{T_2^2}x$
$F_1+F_2=\frac{4m{\pi }^{2}x}{T^2}+\frac{4m{\pi }^{2}x}{T_2^2}$
$\frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}$
$\therefore T=\frac{12}{25}$