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Chapter 9: Problem 67

The ratio of maximum to minimum intensity at a place due to superposition of two waves represented by \(y_{1}=3 \sin (200 t) \mathrm{cm}\) and \(y_{2}=4 \cos (208 t) \mathrm{cm}\) will be (A) \(7: 1\) (B) \(49: 1\) (C) \(4: 3\) (D) \(16: 9\)

Short Answer

Expert verified
The ratio of maximum to minimum intensity due to the superposition of the two waves is equal to \( (A_{max}^2/A_{min}^2) = (7^2/1^2) = 49:1 \). Hence, the correct answer is (B) 49:1.

Step by step solution

01

Express both waves in a common form

Express both waves in sine form using the identity \( \cos x = \sin (x+90掳) \) so you can compare and combine them more easily. Thus, \( y_{2} = 4 \sin (208t+90掳) \).
02

Apply the principle of Superposition

According to the principle of superposition the resultant displacement (y) when two waves superpose is the vector sum of the displacements of the individual waves. In this case, \( y = y_{1}+ y_{2} = 3 \sin (200t) + 4 \sin (208t+90掳) \). It's crucial to remember that the frequencies of the two waves are different.
03

Find out the amplitude of the resultant

The amplitude of the resultant wave is calculated by the formula \( A = \sqrt{ (y_{1})^2 + (y_{2})^2 + 2y_{1}y_{2}\cos \delta } \) where \( \delta \) is the phase difference. In this case, we cannot calculate the exact phase difference because the frequencies of the two waves are different, so \( \delta \) is continuously varying.
04

Analyze amplitude and calculate min and max intensity

The minimum amplitude occurs when the phase difference \( \delta = 180掳 \) and the maximum amplitude is when \( \delta = 0掳 \) or (360掳). So we've two values for A, when Amplitude A is maximum, the resultant intensity \( I_{max} = A_{max}^2 \) and when A is minimum \( I_{min} = A_{min}^2 \). The ratio of maximum and minimum intensity is, therefore, \( I_{max}/I_{min} = A_{max}^2/A_{min}^2 \). From the data given in the problem the minimum and maximum amplitudes can be found to be A_{min} = |3 - 4| = 1 and A_{max} = 3 + 4 = 7.

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