Question

Two waves are simultaneously passing through a string and their equations are :
\(y_1 = A_1 \sin k(x - vt), y_2 = A_2 \sin k(x-vt+x_0)\). Given amplitudes \(A_1 = 12\) mm and \(A_2 = 5\) mm, \(x_0=3.5\) cm and wave number \(k=6.28\) cm\(^{-1}\). The amplitude of resulting wave will be __________ mm.

Hint:

In wave superposition problems, always calculate the phase difference first. Look for special values: if \( \Delta\phi = 2n\pi \) (like 0, 2\(\pi\), ...), interference is constructive and \(A_R = A_1 + A_2\). If \( \Delta\phi = (2n+1)\pi \) (like \(\pi\), 3\(\pi\), ...), interference is destructive and \(A_R = |A_1 - A_2|\). Recognize that 6.28 is a common approximation for \(2\pi\).

Answer 1

Correct Answer: 7

Step 1: Understanding the Question:
We are asked to find the resultant amplitude of two interfering waves traveling in the same direction with a constant phase difference.
Step 2: Key Formula or Approach:
1. The two waves are of the form \(y_1 = A_1 \sin(\phi_1)\) and \(y_2 = A_2 \sin(\phi_2)\). The phase difference is \( \Delta\phi = \phi_2 - \phi_1 \).
2. The resultant amplitude A\(_R\) of the superposition of two waves with amplitudes A\(_1\) and A\(_2\) and a constant phase difference \( \Delta\phi \) is given by:
\[ A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos(\Delta\phi)} \] Step 3: Detailed Explanation:
Given equations:
\(y_1 = A_1 \sin(kx - kvt)\)
\(y_2 = A_2 \sin(kx - kvt + kx_0)\)
The phase of the first wave is \( \phi_1 = kx - kvt \).
The phase of the second wave is \( \phi_2 = kx - kvt + kx_0 \).
The phase difference between the two waves is:
\[ \Delta\phi = \phi_2 - \phi_1 = kx_0 \] Given values:
A\(_1\) = 12 mm
A\(_2\) = 5 mm
x\(_0\) = 3.5 cm
k = 6.28 cm\(^{-1}\)
The value k = 6.28 is a very close approximation of \(2\pi\). So we take \( k = 2\pi \) cm\(^{-1}\).
Now, calculate the phase difference:
\[ \Delta\phi = k x_0 = (2\pi \text{ cm}^{-1}) \times (3.5 \text{ cm}) = 7\pi \text{ radians} \] Now we find the cosine of the phase difference:
\[ \cos(\Delta\phi) = \cos(7\pi) \] Since \( \cos(n\pi) = (-1)^n \) for integer n, we have \( \cos(7\pi) = -1 \).
This means the waves are perfectly out of phase (destructive interference).
Now, use the formula for the resultant amplitude:
\[ A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos(7\pi)} \] \[ A_R = \sqrt{12^2 + 5^2 + 2(12)(5)(-1)} \] \[ A_R = \sqrt{144 + 25 - 120} = \sqrt{169 - 120} = \sqrt{49} = 7 \text{ mm} \] Alternatively, for destructive interference (\(\cos(\Delta\phi) = -1\)), the resultant amplitude is simply the difference between the individual amplitudes:
\[ A_R = |A_1 - A_2| = |12 - 5| = 7 \text{ mm} \] Step 4: Final Answer:
The amplitude of the resulting wave will be 7 mm.