First slide

Applications of SHM

Question

A body performs SHM along the straight line segment ABCDE with c as the mid point of segment AE (A and E are the extreme position for the SHM). Its kinetic energies at B and D are each one fourth of its maximum value. The length of segment AE is 2R, if the distance between B and, D is found to be $\sqrt{x} R$ . The value of x is __________.

Moderate

Solution

When the particle crosses point D, its speed is half of the maximum speed. Given that amplitude is 2R
$\begin{array}{l} \Rightarrow v = \frac{v_{max}}{R} \sqrt{R^{2} - x^{2}} \\ or \frac{v_{max}}{2} = \frac{v_{max}}{R} \sqrt{R^{2} - x^{2}} \\ or x = \frac{\sqrt{3}}{2} R \\ \Rightarrow Distance BD = 2 x = \sqrt{3} R \\ Hence x = 3 \end{array}$

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