Two wave sources S1 and S2 are L = 2 m apart as shown in the figure below. They both emit coherent harmonic waves of equal amplitude A0, frequency and wavelength. Find:
- The amplitude of the resultant wave at point P.
- The intensity of the resultant wave at the same point P.
Datos: speed of sound v = 340 m/s; frequency ν = 2000 Hz; α = 300; intensity emitted by each source I0.
Ad blocker detected
Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!
Solution:
In this problem the interference of the two waves is caused by the difference in distance traveled by the two waves, also known as path difference. The amplitude of the resultant wave is given by:
where k is the wavenumber that, in terms of the wavelength λ is:
On the other hand, x1 and x2 are the distances traveled by both waves from their source to point P.
Since both the distance L between the two sources and the angle α are known, we can determine both distances using some trigonometry.
And substituting into the expression for A’ we get:
Now, substituting the numerical values:
The expression gives a negative result but the amplitude of a wave is always a positive number.
Ad blocker detected
Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!
The intensity of a harmonic wave of amplitude A0 is given by:
This expression gives the intensity emitted by each source.
Since the amplitude A’ of the resultant wave at point P is known, we can determine the total intensity I at this same point P using the same expression written as in terms of I0:
