Question

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

• A. 0.72 sec
• B. 0.64 sec
• C. 0.48 sec
• D. 0.36 sec

Answer 1

The correct option is C

0.48 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 = $F_1+F_2$

$\Rightarrow m{\omega }^{2}y=m{\omega }_1^{2}y+m{\omega }^{2}y$

$\Rightarrow {\omega }^2={\omega }_1^2+{\omega }_2^2\Rightarrow {\left(\frac{2\pi }{T}\right)}^2={\left(\frac{2\pi }{T_1}\right)}^2={\left(\frac{2\pi }{T_2}\right)}^2$

$\Rightarrow T=\sqrt{\frac{T_1^2{T}_2^2}{T_1^2+T_2^2}}=\sqrt{\frac{{\left(\frac{4}{5}\right)}^2{\left(\frac{3}{5}\right)}^2}{{\left(\frac{4}{5}\right)}^2+{\left(\frac{3}{5}\right)}^2}}$ = 0.48 sec