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)

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 $F=F_1+F_2$
$\Rightarrow m{\omega }^{2}y=m{\omega }_1^{2}y+m{\omega }_2^{2}y$
$\Rightarrow {\omega }^2={\omega }_1^2+{\omega }_2^2$
$\Rightarrow {\left[\frac{2\pi }{T}\right]}^2={\left[\frac{2\pi }{T_1}\right]}^2+{\left[\frac{2\pi }{T_2}\right]}^2$
$\Rightarrow T=\sqrt{\frac{T_1^2\times T_2^2}{T_1^2+T_2^2}}$
$=\sqrt{\frac{{\left(\frac{4}{5}\right)}^2{\left(\frac{3}{5}\right)}^2}{{\left(\frac{4}{5}\right)}^2+{\left(\frac{3}{5}\right)}^2}}$
$=\frac{12}{25}s$