On the superposition of two harmonic oscillat | vedclass

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(A) The resultant amplitude $A$ of two superposed harmonic oscillations with amplitudes $a_1$ and $a_2$ is given by $A = \sqrt{a_1^2 + a_2^2 + 2a_1 a_

Vedclass · Accepted Answer

$120$

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(A) The resultant amplitude $A$ of two superposed harmonic oscillations with amplitudes $a_1$ and $a_2$ is given by $A = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos(\phi_1 - \phi_2)}$.Given that $a_1 = a_2 = a$ and the resultant amplitude is also $a$,we have:$a = \sqrt{a^2 + a^2 + 2a^2 \cos(\phi_1 - \phi_2)}$Squaring both sides:$a^2 = 2a^2 + 2a^2 \cos(\phi_1 - \phi_2)$Dividing by $a^2$:$1 = 2 + 2 \cos(\phi_1 - \phi_2)$$2 \cos(\phi_1 - \phi_2) = -1$$\cos(\phi_1 - \phi_2) = -1/2$Therefore,$\phi_1 - \phi_2 = \cos^{-1}(-1/2) = 120^o$.

On the superposition of two harmonic oscillations represented by $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$,a resulting oscillation with the same time period and amplitude is obtained. The value of $\phi_1 - \phi_2$ is .... $^o$

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