Question

Due to some force $F_1$, a body oscillates with period $\frac{4}{5}\text{sec}$, and due to other force $F_2$, oscillates with period $\frac{3}{5}\text{sec}$. If both forces act simultaneously, the new period will be :

• A. $0.72\text{sec}$
• B. $0.64\text{sec}$
• C. $0.48\text{sec}$
• D. $0.36\text{sec}$

Answer 1

The correct option is C $0.48\text{sec}$

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 $F=F_1+F_2$
$\Rightarrow m{\omega }^{2}y=m{\omega }_1^{2}y+m{\omega }_2^{2}y$
$\Rightarrow \omega ={\omega }_1^2+{\omega }_2^2$
$\Rightarrow {\left(\frac{2\pi }{T}\right)}^2={\left(\frac{2\pi }{T_1}\right)}^2+{\left(\frac{2\pi }{T_2}\right)}^2$
$\Rightarrow T=\sqrt{\frac{T_1^2{T}_2^2}{T_1^2+T_2^2}}$
$=\sqrt{\frac{{\left(\frac{4}{5}\right)}^2{\left(\frac{3}{5}\right)}^2}{{\left(\frac{4}{5}\right)}^2+{\left(\frac{3}{5}\right)}^2}}=0.48\text{sec}$