Physics : Waves - Solved Example Problems for Waves Superposition Principle

Consider two sources A and B as shown in the figure below. Let the two sources emit simple harmonic waves of same frequency but of different amplitudes, and both are in phase (same phase). Let O be any point equidistant from A and B as shown in the figure. Calculate the intensity at points O, Y and X. (X and Yâ€™ are not equidistant from A & B)

The distance between OA and OB are the same and hence, the waves starting from A and B reach O after covering equal distances (equal path lengths). Thus, the path difference between two waves at O is zero.

*OA *âˆ’* OB *= 0

Since the waves are in the same phase, at the point O, the phase difference between two waves is also zero. Thus, the resultant intensity at the point O is maximum.

Consider a point Y, such that the path difference between two waves is *Î»*. Then the phase difference at Y is

Therefore, at the point Y, the two waves from A and B are in phase, hence, the intensity will be maximum.

Consider a point X, and let the path difference the between two waves be Î»/2.

Then the phase difference at X is

Therefore, at the point X, the waves meet and are in out of phase, Hence, due to destructive interference, the intensity will be minimum.

Two speakers C and E are placed 5 m apart and are driven by the same source. Let a man stand at A which is 10 m away from the mid point O of C and E. The man walks towards the point O which is at 1 m (parallel to OC) as shown in the figure. He receives the first minimum in sound intensity at B. Then calculate the frequency of the source.

(Assume speed of sound = 343 m s-1)

The first minimum occurs when the two waves reaching the point B are 180Â° (out of phase). The path difference âˆ†x = Î»/2.

In order to calculate the path difference, we have to find the path lengths *x*1 and *x*2.

In a right triangle BDC,

The path difference âˆ†*x* = *x*2 âˆ’ *x*1 = 10.6 mâˆ’10.1 m = 0.5 m. Required that this path difference

Consider two sound waves with wavelengths 5â€†*m* and 6â€†*m*. If these two waves propagate in a gas with velocity 330â€†*ms*-1. Calculate the number of beats per second.

Given *Î»*1 = 5*m* and *Î»*2 = 6*m*

Velocity of sound waves in a gas is *v *= 330* ms*-1

The relation between wavelength and velocity is *v* = *Î»f* => *f** *= v/Î»

The number of beats per second is

| *f*1 âˆ’ *f*2| = |66 âˆ’ 55| = 11 beats per sec

Two vibrating tuning forks produce waves whose equation is given by *y*1 = 5 sin(240*Ï€ t*) and *y*2 = 4 sin(244*Ï€t*). Compute the number of beats per second.

Given *y*1 = 5 sin(240*Ï€ t*) and *y*2 = 4 sin(244*Ï€t*)

Comparing with *y* = *A* sin(2*Ï€ f*1*t*), we get

2*Ï€f*1 = 240*Ï€* â‡’ *f*1 = 120*Hz*

2*Ï€f*2 = 244*Ï€* â‡’ *f*2 = 122*Hz*

The number of beats produced is | *f*1 âˆ’ *f*2| = |120 âˆ’ 122| = |âˆ’ 2|=2 beats per sec

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