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
Text Solution
Verified by ExpertsC
Under the influence of one force $\mathbf{F}_1 = m \omega_1^2 y$ and under the action of another force, $\mathbf{F}_2 = m \omega_2^2 y$ .
Under the action of both the forces $\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2$
$\Rightarrow$ $m \omega^{2} y = m \omega_{1}^{2} y + m \omega^{2} y$
$\Rightarrow \omega_1^2 + \omega_2^2 \Rightarrow \left(\frac{2\pi}{T}\right)^2 = \left(\frac{2\pi}{T_1}\right)^2 + \left(\frac{2\pi}{T_2}\right)^2$ $\Rightarrow T = \sqrt{\frac{T_1^2 T_2^2}{T_1^2 + T_2^2}} = \sqrt{\frac{\left(\frac{4}{5}\right)^2 \left(\frac{3}{5}\right)^2}{\left(\frac{4}{5}\right)^2 + \left(\frac{3}{5}\right)^2}} = 0.48 \mathrm{sec}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems