problem about interference - waves

Question

You have two vibrating objects in an infinitely large pool. The distance between them is 6.0 m. Their frequency of vibration is 2.0 Hz and the wave speed is 4.0 m/s. The vibrations are sinusoidal. Find a few points where they are in phase and out of phase.

Answer 1

The easiest place to look is along a line between the sources.  Choose any point between the sources.  It will be some distance d from one source, and a distance (6-d) from the other source .  The expression for the value of phase φ along some point of a sinusoidal wave is:

 

φ = kx-ωt 

 

where

 

k = (angular) wavenumber = 2π/λ, where λ=wavelength

ω = angular frequency = 2πf, where f = (linear) frequency in Hz

 

So, at a point a distance d from one source, along a line between the two sources, the wave from each source has a phase

 

φ₁ = kd - ωt   (I am taking "Source 1" to be the one that this point is a distance d from)

φ₂ = k(6-d)-ωt

 

Since we are looking at the same time at both waves, and they have the same frequency, the ωt terms are identical.

 

A sine wave repeats at intervals of 2π radians, so if we two equal sine waves (in frequency and wavelength) side-by-side, and then we slide one in either direction by 2π radians (or any integer multiple of 2π radians) with respect to the other, they will still be in phase.  Therefore, two sine waves will be in phase if the magnitude of the phase difference between them at a given point is equal to either 0 or an integer multiple of 2π.  Thus, we need:

 

|φ₂-φ₁|= |(kd-ωt)-(k(6-d)-ωt)|= |k(d-(6-d))| =| k(2d-6)| = |((2π)/λ)(2d-6)| =  2πm

 

The factors of 2π on each side cancel out, leaving:

 

|(2d-6)/λ| = m,  with m = 0,1,2,3...

 

The value for the wavelength can be calculated from the given quantities, using:

 

v=fλ --> 4 m/s = (2 Hz)(λ) --> λ = 2 m (as Hilton showed)

 

This means, for the two waves being in phase between the sources, we have:

 

|(2d-6)/2| = m  (m= 1,2,3...) --> |d-3| = m

 

This breaks up through the magnitude (absolute value) symbol, into

 

d - 3 = m

-(d-3) = 3 - d = m

 

Now, for this setup to be valid, the value of d cannot exceed 6 m (otherwise, we are no longer between the sources, and the geometry that derived the previous expression no longer holds).  So, we need to start at m = 1, and continue solving for d for each allowed value of m, until d exceeds 6 m.

 

This means the allowed values for d, for in phase, are:

 

d = 1 m, 2 m, 3 m, 4 m, 5 m 

 

Technically, anywhere they are not in phase, they are "out of phase," but, if they mean 180^o or π radians out of phase, we just have to change the condition of the magnitude of the phase difference from being 0 or an integer multiple of 2π to being an odd-integer multiple of π.  This because, if you have two sine waves in phase, and then slide one of them either way by one-half of 2π radians (or one-half of 4π or 6π, and so on, radians), you end up with the waves being π radians out of phase.  So this means

 

|d-3| = (m+1/2)  with m=0,1,2,3...

 

Again, this breaks down into two expressions:

 

d-3 = (m+1/2)

3-d = (m+1/2)

 

Starting at m = 0, we cycle through the allowed values of m and note the solutions for d that do not exceed 6 m.  For this case, then, we have allowed solutions for out-of-phase d of:

 

d = 0.5, 1.5, 2.5, 3.5, 4.5, 5.5 m

 

 

If we think of a string tied between the two sources, in a resonance at the given frequency and wavelength, the in-phase positions correspond to the nodes of the string, and the out-of-phase positions correspond to the antinodes.

 

 

 

 

Answer 2

We are given the following

v = 4 m/s

f = 2 Hz

x = 6 m

 

From this, λ = v/f = 4/2 = 2 m

This means that there are three (x/λ = 3) waves between the two vibrating objects.

Assume that the first vibrating object is at x =0,  and the second is at x = 6 m.

Points that are 1λ, or 2λ or 3λ apart are in phase; points that are 0.5λ, or 1.5λ or 2.5λ apart are out of phase.

If you cannot see this, sketch the sine graph with the above properties.