Two particles of mass m are constrained to mo | vedclass

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(B) Let $x$ be the distance of each particle from the vertex (meeting point) of the rails. The distance between the two particles is $d = 2x \sin 	he

Vedclass · Accepted Answer

$\sqrt{\frac{2k}{m}} \sin 	heta$

Vedclass · Answer

(B) Let $x$ be the distance of each particle from the vertex (meeting point) of the rails. The distance between the two particles is $d = 2x \sin 	heta$.If the particles are displaced by a small amount $\Delta x$ from their equilibrium position $x_0$,the change in the length of the spring is $\Delta d = 2 \Delta x \sin 	heta$.The potential energy of the spring is $U = \frac{1}{2} k (\Delta d)^2 = \frac{1}{2} k (2 \Delta x \sin 	heta)^2 = 2 k \sin^2 	heta (\Delta x)^2$.The kinetic energy of the two particles is $K = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2 = m (\Delta \dot{x})^2$.Comparing this with the standard form of energy for a simple harmonic oscillator,$E = \frac{1}{2} k_{eff} x^2 + \frac{1}{2} m_{eff} \dot{x}^2$,we identify $k_{eff} = 4 k \sin^2 	heta$ and $m_{eff} = 2m$.The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k_{eff}}{m_{eff}}} = \sqrt{\frac{4 k \sin^2 	heta}{2m}} = \sqrt{\frac{2k}{m}} \sin 	heta$.

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