Question

A particle of mass M, is executing simple harmonic oscillations of amplitude A. The maximum energy, of this oscillating particle, is E₀.

The time period, T, of this oscillating particle, and its velocity v_x , when it is displaced a distance \(x(=\frac{A\sqrt{3}}{2})\) from its mean position, can be expressed, in terms of E₀ , as

(1) \(T = (\frac{2\pi^2 MA^2}{E_0})^{1/2} ; \,v_x =(\frac{E_0}{2M})^{1/2}\)

(2) \(T = (\frac{\pi^2 MA^2}{E_0})^{1/2} ; \,v_x =(\frac{E_0}{2M})^{1/2}\)

(3) \(T = (\frac{2\pi^2 MA^2}{E_0})^{1/2} ; \,v_x =(\frac{2E_0}{2M})^{1/2}\)

(4) \(T = (\frac{\pi^2 MA^2}{2E_0})^{1/2} ; \,v_x =(\frac{E_0}{2M})^{1/2}\)

Answer 1

 (1) \(T = (\frac{2\pi^2 MA^2}{E_0})^{1/2} ; \,v_x =(\frac{E_0}{2M})^{1/2}\)

The maximum energy of a particle, executing SHM, is given by