Question

A box 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 S.H.M 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 will be

• A. $\frac{12}{25}$
• B. $\frac{24}{25}$
• C. $\frac{35}{24}$
• D. $\frac{15}{12}$

Answer 1

The correct option is A $\frac{12}{25}$

Let the body be displaced by a distance x. If the restoring force is $F_1$, then the angular frequency of the resulting simple harmonic motion is given by
${\omega }_1^2=\frac{K}{m}=\frac{Kx}{mx}=\frac{F_1}{mx}$ .....(i)
where m is the mass of the body.
For force $F_2$, we have, ${\omega }_2^2=\frac{F_2}{mx}$ ..... (ii)
Now, ${\omega }^2=\frac{F_1+F_2}{mx}$ ..... (iii)
From (i), (ii) and (iii), we get
${\omega }^2={\omega }_1^2+{\omega }_2^2$
$\Rightarrow {\left(\frac{2\pi }{T}\right)}^2={\left(\frac{2\pi }{T_1}\right)}^2+{\left(\frac{2\pi }{T_2}\right)}^2$
$\Rightarrow \frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}=\frac{1}{{\left(\frac{4}{5}\right)}^2}+\frac{1}{{\left(\frac{3}{5}\right)}^2}$
$\Rightarrow \frac{1}{T^2}=\frac{25}{16}+\frac{25}{9}=\frac{25\times 25}{16\times 9}=\frac{25\times 25}{144}$
$\Rightarrow T=\sqrt{\frac{144}{25\times 25}}=\frac{12}{25}s$.
Hence, the correct choice is (a).