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(B) The resultant amplitude $A_{	ext{res}}$ of two simple harmonic motions with amplitudes $A_1$ and $A_2$ and phase difference $\delta$ is given by

Vedclass · Accepted Answer

$\delta=\frac{2 \pi}{3}$

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(B) The resultant amplitude $A_{	ext{res}}$ of two simple harmonic motions with amplitudes $A_1$ and $A_2$ and phase difference $\delta$ is given by the formula: $A_{	ext{res}}^2 = A_1^2 + A_2^2 + 2A_1A_2 \cos \delta$. Given that $A_1 = A_2 = A$ and the resultant amplitude $A_{	ext{res}} = A$, we substitute these values into the equation: $A^2 = A^2 + A^2 + 2A^2 \cos \delta$. $A^2 = 2A^2 + 2A^2 \cos \delta$. $A^2 - 2A^2 = 2A^2 \cos \delta$. $-A^2 = 2A^2 \cos \delta$. $\cos \delta = -\frac{1}{2}$. Since $\cos \delta = -\frac{1}{2}$, the phase difference $\delta = 120^{\circ}$ or $\delta = \frac{2\pi}{3}$ radians.

$A$ particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motion, the phase difference $(\delta)$ between the two motions is

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