Question

The value of maximum amplitude produced due to interference of two waves given by \(Y_1 = 4\sin(\omega t)\) and \(Y_2 = 3\cos(\omega t)\) is

• A. 7
• B. 5
• C. 1
• D. 25

Hint:

When two waves are given in sine and cosine form with the same frequency, they have a phase difference of \(\pi/2\). The resultant amplitude can be found quickly using the Pythagorean theorem: \(A_R = \sqrt{A_{sin}^2 + A_{cos}^2}\). In this case, \(A_R = \sqrt{4^2 + 3^2} = 5\), which is a common Pythagorean triple (3-4-5).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When two or more waves superpose, the resultant displacement is the vector sum of the individual displacements. The amplitude of the resultant wave depends on the amplitudes of the individual waves and the phase difference between them. The question asks for the amplitude of the resultant wave formed by the superposition of \(Y_1\) and \(Y_2\).
Step 2: Key Formula or Approach:
The two waves are given by: \(Y_1 = 4\sin(\omega t)\) \(Y_2 = 3\cos(\omega t) = 3\sin(\omega t + \frac{\pi}{2})\) The amplitudes are \(A_1 = 4\) and \(A_2 = 3\). The phase difference between them is \(\phi = \frac{\pi}{2}\) (or 90 degrees).
The amplitude of the resultant wave (\(A_R\)) is given by the formula for vector addition: \[ A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi} \] Step 3: Detailed Explanation:
Substitute the values into the formula: \(A_1 = 4\), \(A_2 = 3\), and \(\phi = \frac{\pi}{2}\). \[ A_R = \sqrt{4^2 + 3^2 + 2(4)(3)\cos(\frac{\pi}{2})} \] Since \(\cos(\frac{\pi}{2}) = 0\), the last term becomes zero. \[ A_R = \sqrt{4^2 + 3^2 + 0} \] \[ A_R = \sqrt{16 + 9} = \sqrt{25} \] \[ A_R = 5 \] The resultant wave has an amplitude of 5. The term "maximum amplitude" in the question refers to the amplitude of this resulting wave.
Step 4: Final Answer:
The amplitude of the wave produced by the interference of the two given waves is 5. Therefore, option (B) is correct.