Q. Due to some force $F_{1}$ a body oscillates with period $4/5\ s$ and due to other force $F_{2}$ oscillates with period $3/5\ s.$ If both forces act simultaneously, the new period will be

A) $0.72\ s$
B) $0.64\ s$
C) $0.48\ s$
D) $0.36\ s$
Explanation

(c) 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 $F = F_{1} + F_{2}$
$\Rightarrow m\omega^{2}y = m\omega_{1}^{2}y + m\omega_{2}^{2}y$
$\Rightarrow \omega^{2} = \omega_{1}^{2} + \omega_{2}^{2} \Rightarrow \left( \frac{2\pi}{T} \right)^{2} = \left( \frac{2\pi}{T_{1}} \right)^{2} + \left( \frac{2\pi}{T_{2}} \right)^{2}$
$\Rightarrow T = \sqrt{\frac{T_{1}^{2}T_{2}^{2}}{T_{1}^{2} + T_{2}^{2}}} = \sqrt{\frac{\left( \frac{4}{5} \right)^{2}\left( \frac{3}{5} \right)^{2}}{\left( \frac{4}{5} \right)^{2} + \left( \frac{3}{5} \right)^{2}}} = 0.48s$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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