The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, $y_1(x,t) = 4 \sin(kx – \omega t)$ and $y_2(x,t) = 2 \sin(kx – \omega t + \frac{2\pi}{3})$, are:
Options: A) $[\sqrt{3}, \frac{\pi}{6}]$, B) $[2\sqrt{3}, \frac{\pi}{6}]$, C) $[6, \frac{2\pi}{3}]$, D) $[6, \frac{\pi}{3}]$
Options: A) $[\sqrt{3}, \frac{\pi}{6}]$, B) $[2\sqrt{3}, \frac{\pi}{6}]$, C) $[6, \frac{2\pi}{3}]$, D) $[6, \frac{\pi}{3}]$
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Solution Steps
1
Extract Wave Parameters
From the given equations $y_1 = 4 \sin(kx – \omega t)$ and $y_2 = 2 \sin(kx – \omega t + \frac{2\pi}{3})$:
Amplitude of first wave, $A_1 = 4$
Amplitude of second wave, $A_2 = 2$
Phase difference between them, $\phi = \frac{2\pi}{3}$ ($120^\circ$)
2
Calculate Resultant Amplitude (A)
Using the phasor addition formula:
$$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$$
$$A = \sqrt{4^2 + 2^2 + 2(4)(2) \cos\left(\frac{2\pi}{3}\right)}$$
$$A = \sqrt{16 + 4 + 16\left(-\frac{1}{2}\right)} = \sqrt{20 – 8} = \sqrt{12}$$
$$A = 2\sqrt{3}$$
3
Find Resultant Phase Angle ($\alpha$)
The resultant phase $\alpha$ relative to the first wave is given by:
$$\tan \alpha = \frac{A_2 \sin \phi}{A_1 + A_2 \cos \phi}$$
$$\tan \alpha = \frac{2 \sin(120^\circ)}{4 + 2 \cos(120^\circ)} = \frac{2 \cdot \frac{\sqrt{3}}{2}}{4 + 2 \cdot (-\frac{1}{2})}$$
$$\tan \alpha = \frac{\sqrt{3}}{4 – 1} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$
4
Determine the Final Phase
Since $\tan \alpha = \frac{1}{\sqrt{3}}$, the phase angle is:
$$\alpha = 30^\circ = \frac{\pi}{6}$$
5
Combine Results
The resultant amplitude is $2\sqrt{3}$ and the resultant phase is $\frac{\pi}{6}$.
Final Result: [2√3, π/6]
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Theory
1. Superposition of Harmonic Waves
When two harmonic waves of the same frequency and wavelength travel in the same direction, they combine to form a single resultant wave. The displacement of the resultant wave is the sum of the individual displacements. Mathematically, $y_{res} = y_1 + y_2$. This resultant wave maintains the same frequency ($\omega$) and wave number ($k$) but features a new amplitude and a new phase constant. This phenomenon is the basis for interference in optics and acoustics.
2. Phasor Method for Waves
Calculations involving trigonometric sums can be simplified by representing waves as rotating vectors called phasors. The length of the phasor represents the amplitude, and its angle with the horizontal axis represents the phase. Adding two waves is equivalent to adding their corresponding vectors using the parallelogram law. This leads to the resultant amplitude formula $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$, which is identical to the vector addition formula.
3. Resultant Phase Angle
The resultant phase $\alpha$ determines the ‘shift’ of the new wave relative to the original reference wave ($y_1$). It is derived from the geometry of the phasor diagram. The formula $\tan \alpha = (A_2 \sin \phi) / (A_1 + A_2 \cos \phi)$ works for any phase difference $\phi$. It is important to note that if $\phi$ is in the second or third quadrant, one must be careful with the signs of $\sin$ and $\cos$ to ensure the correct quadrant for $\alpha$.
4. Trigonometric Constants in Wave Problems
JEE problems frequently use standard angles like $\frac{\pi}{3} (60^\circ)$, $\frac{2\pi}{3} (120^\circ)$, and $\pi (180^\circ)$ to test the student’s speed with unit circle values. Knowing that $\cos(120^\circ) = -0.5$ and $\sin(120^\circ) = \sqrt{3}/2$ is essential for quick computation. These specific values often lead to simplified results like $\sqrt{3}$ or $2\sqrt{3}$, which are common in multiple-choice options.
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FAQs
1
What if the waves were moving in opposite directions?
If they moved in opposite directions, they would form a standing wave instead of a travelling resultant wave, and the amplitude would vary with position (x).
2
Can the amplitude ever be 6?
Yes, if the phase difference $\phi = 0$ (constructive interference), the amplitude would be $A_1 + A_2 = 4 + 2 = 6$.
3
Can the amplitude be 2?
Yes, if the phase difference $\phi = \pi$ (destructive interference), the amplitude would be $|A_1 – A_2| = |4 – 2| = 2$.
4
Why did we use tan to find the phase?
The tangent ratio comes from the vector component breakdown (Vertical/Horizontal) of the resultant phasor in the coordinate plane.
5
What is the significance of the 2π/3 phase?
It represents a time or space delay of the second wave by one-third of its total cycle ($120^\circ$).
6
Is the energy of the resultant wave conserved?
Energy is proportional to amplitude squared. The energy of the resultant wave ($12$) is less than the sum of individual energies ($16 + 4 = 20$) because of partial destructive interference.
7
How do I convert degrees to radians quickly?
Multiply by $\pi/180$. For example, $30^\circ \times (\pi/180) = \pi/6$.
8
Does the ‘x’ or ‘t’ affect the amplitude calculation?
No, the amplitude is a constant peak value; $x$ and $t$ only determine the instantaneous displacement.
9
Is this wave transversal or longitudinal?
The mathematical form $\sin(kx – \omega t)$ can represent both, though it is most commonly used for transverse string waves or light waves.
10
What if there were three waves?
You would add their phasors tip-to-tail or sum their x and y components separately and then find the resultant magnitude.
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