Question

Consider two waves, which are given by $y_1(x, t) = A \sin(kx - \omega t)$ and $y_2(x, t) = \sqrt{3}A \cos(kx - \omega t)$, where $k$ is the wave number and $\omega$ is the angular frequency. The amplitude of the resultant waveform obtained by the superposition of the two waves is $A_s$ and its phase difference with $y_1$ is $\phi_s$. What are $A_s$ and $\phi_s$?

• A. $A_s = 2A$ and $\phi_s = \frac{\pi}{3}$
• B. $A_s = 2A$ and $\phi_s = \frac{\pi}{6}$
• C. $A_s = \frac{A}{2}$ and $\phi_s = \frac{\pi}{3}$
• D. $A_s = \frac{A}{2}$ and $\phi_s = \frac{\pi}{6}$

Hint:

Using phasors, the two waves can be represented as vectors: $\vec{A}_1$ along the x-axis with magnitude $A$, and $\vec{A}_2$ along the y-axis (since cosine leads sine by $\pi/2$) with magnitude $\sqrt{3}A$.
The magnitude of the resultant vector is $A_s = \sqrt{A^2 + (\sqrt{3}A)^2} = 2A$.
The phase angle $\phi_s$ with respect to the x-axis is $\tan^{-1}\left(\frac{\sqrt{3}A}{A}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question asks for the resultant amplitude and relative phase of two superimposed harmonic waves.
Step 2: Key Formulas and Approach:
1. Represent the cosine wave in terms of a sine wave to easily find the phase difference:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
2. Alternatively, use trigonometric expansion of the linear combination:
\[ X \sin \theta + Y \cos \theta = \sqrt{X^2 + Y^2} \sin(\theta + \phi_s) \]
where $\tan \phi_s = \frac{Y}{X}$.
Step 3: Detailed Explanation:

Let us write the total displacement $y = y_1 + y_2$:
\[ y = A \sin(kx - \omega t) + \sqrt{3}A \cos(kx - \omega t) \]

We can rewrite the sum by multiplying and dividing the right-hand side by $\sqrt{A^2 + (\sqrt{3}A)^2} = \sqrt{A^2 + 3A^2} = 2A$:
\[ y = 2A \left[ \frac{1}{2} \sin(kx - \omega t) + \frac{\sqrt{3}}{2} \cos(kx - \omega t) \right] \]

We know that:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \quad \text{and} \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

Substituting these values into the bracketed term:
\[ y = 2A \left[ \sin(kx - \omega t) \cos\left(\frac{\pi}{3}\right) + \cos(kx - \omega t) \sin\left(\frac{\pi}{3}\right) \right] \]

Using the trigonometric identity $\sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta$, we can simplify the expression:
\[ y = 2A \sin\left(kx - \omega t + \frac{\pi}{3}\right) \]

Comparing this resultant wave with the standard form $y = A_s \sin(kx - \omega t + \phi_s)$:

The resultant amplitude is $A_s = 2A$.

The phase difference with respect to $y_1(x, t) = A \sin(kx - \omega t)$ is $\phi_s = \frac{\pi}{3}$.


Step 4: Final Answer:
The amplitude $A_s$ is $2A$ and the phase difference $\phi_s$ is $\frac{\pi}{3}$, which corresponds to Option (A).