Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The displacement of a particle executing simple harmonic motion is given by $y = A_0 + A \sin \omega t + B \cos \omega t$ . Then the amplitude of its oscillation is given by:

$A + B$

$A_0 + \sqrt{A^2 + B^2}$

$\sqrt{A^2 + B^2}$

$\sqrt{A_0^2 + (A+B)^2}$

Step-by-Step Solution

Analyze the Equation: The given displacement equation is $y = A_0 + A \sin \omega t + B \cos \omega t$ .
Identify Components:

The term $A_0$ is a constant, representing the shift of the equilibrium position from the origin. It does not affect the amplitude of the oscillation.
The oscillating part consists of two simple harmonic motions: $y_1 = A \sin \omega t$ and $y_2 = B \cos \omega t$ .

Determine Phase Difference: We know that $\cos \omega t = \sin(\omega t + \frac{\pi}{2})$ . Therefore, the second wave leads the first by a phase angle of $\phi = \frac{\pi}{2}$ .
Apply Superposition Principle: The resultant amplitude $R$ of two SHMs with amplitudes $A$ and $B$ and phase difference $\phi$ is given by the formula : $R = \sqrt{A^2 + B^2 + 2AB \cos \phi}$
Calculate Amplitude: Substituting $\phi = \frac{\pi}{2}$ : $R = \sqrt{A^2 + B^2 + 2AB \cos(90^\circ)}$ $R = \sqrt{A^2 + B^2 + 0}$ $R = \sqrt{A^2 + B^2}$

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