The displacement of a particle executing simple harmonic motion is given by $y = A_0 + A \sin \omega t + B \co

Question

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:

Accepted Answer

1. **Analyze the Equation:** The given displacement equation is $y = A_0 + A \sin \omega t + B \cos \omega t$. 
2. **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 $y_2 = B \cos \omega t$.
3. **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 $\phi = \frac{\pi}{2}$.
4. **Apply Superposition Principle:** The resultant amplitude $R$ of two SHMs with amplitudes $A$ and $B$ and phase difference $\phi$ is given by the formula :
   $$ R = \sqrt{A^2 + B^2 + 2AB \cos \phi} $$
5. **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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