How Do Sound Waves Interfere and How Does Light Travel in a Medium?

[dekoi]

1.) Two speakers are on a wall 2m apart. A listerner stands 3.00m from wall in front of one of the speakers. An oscillator runs both speakers at 300Hz.
a.) What is the phase difference between two waves when they reach observer?
b.) What is the frequency closest to 300Hz to which the oscillator may be adjusted such that the observer hears minimal sound?

I began by finding the path difference between the two waves. One is assumed to be on the ground, so its path is 3.0 m long. The other is ~3.5 (sqrt(13)). For part (a) path difference is 0.61m. $\Delta r = \frac{\phi}{2\pi}{\lambda}$ and $\lambda = \frac{v}{f}$, hence i solved for $\phi$, and it turned out to be 3.35 rad. Can anyone verify this, and most importantly explain why i even did this? Any explanation would be great.
Part B asks me for the frequency when there is minimal sound. Minimal sound occurs when there is destructive interference. So according to my notes i equated the angle to $(2n+1)\pi$. Is this correct? What should i solve for and why?


2.) Two speakers are driven in phase by a common oscillator at 800Hz and face each other at distance of 1.25m. Locate the points along a line joining the two speakers where relative minima of sound pressure amplitude would be expected.

Once again, I am solving for the points where there is destructive interference. I could solve for wavelength via $\frac{v}{f}$. Reload this page in a moment.. Thus wavelength equals 2.33m. Then i used the equation $d=\frac{(2n+1)\lambda}{2}$. Is this correct? What do i solve for? Where do i go from there?


3.) Two sinusoidal waves combine in medium are described by:
$y = 3sin\pi(x + 0.6t)$
and
$y = 3sin\pi(x - 0.6t)$
Determine max transverse poistion of an element in medium at 0.250 cm.

Any beginning suggestions would be great. I suppose i could find the sum of the two waves (the superposition), but I am not sure where to go from there.


4.) A cube (n = 1.59) has a small air bubble inside. When a 1.90cm coin is placed over the bubble on the outside of cube, the bubble cannon be seen by looking down into the cube at any angle. However, when a 1.75cm coin is placed directly over it, the bubble can be seen by looking down into the cube. What is the range of the possible depths of the air bubble beneath the surface?

I attempted to draw the diagram for this, but couldn't get a clear understanding of the situation. If anyone could draw me a sample diagram, i would be very grateful.

 

[Andrew Mason]

1.) Two speakers are on a wall 2m apart. A listerner stands 3.00m from wall in front of one of the speakers. An oscillator runs both speakers at 300Hz.
a.) What is the phase difference between two waves when they reach observer?
b.) What is the frequency closest to 300Hz to which the oscillator may be adjusted such that the observer hears minimal sound?
I began by finding the path difference between the two waves. One is assumed to be on the ground, so its path is 3.0 m long. The other is ~3.5 (sqrt(13)). For part (a) path difference is 0.61m. $\Delta r = \frac{\phi}{2\pi}{\lambda}$ and $\lambda = \frac{v}{f}$, hence i solved for $\phi$, and it turned out to be 3.35 rad. Can anyone verify this, and most importantly explain why i even did this? Any explanation would be great.

The answer is right. The difference in path length means that the two sound waves will not be identical at that point. The phase difference is the number of wave lenghths difference divided by $2\pi$.

Part B asks me for the frequency when there is minimal sound. Minimal sound occurs when there is destructive interference. So according to my notes i equated the angle to $(2n+1)\pi$. Is this correct? What should i solve for and why?

When two waves interfere the amplitudes of the two sound waves simply add together. If there is a half wavelength or at odd multiples of $\pi$ phase difference, the amplitudes sum to zero (one at a peak the other at a trough) and you have a node.

2.) Two speakers are driven in phase by a common oscillator at 800Hz and face each other at distance of 1.25m. Locate the points along a line joining the two speakers where relative minima of sound pressure amplitude would be expected.

Once again, I am solving for the points where there is destructive interference. I could solve for wavelength via $\frac{v}{f}$. Reload this page in a moment.. Thus wavelength equals 2.33m. Then i used the equation $d=\frac{(2n+1)\lambda}{2}$. Is this correct? What do i solve for? Where do i go from there?

What are you using for the speed of sound? It should be about 344 m/s assuming room temperature. I get .43 m for the wavelength. So nodes will occur where the differences between the two speakers are odd multiples of half wavelengths. Work out the expression for the difference in distances at a point x between the speakers and use your formula $d=\frac{(2n+1)\lambda}{2}$
AM

 

[quicknote]

Also, some guidance on how to approach this problem would be helpful.

1.) The phase difference between the two waves when they reach the observer can be calculated using the path difference, which is the difference in distance traveled by the two waves to reach the observer. In this case, the path difference is 0.61m. Using the equation \Delta r = \frac{\phi}{2\pi}{\lambda} and \lambda = \frac{v}{f}, we can solve for the phase difference, which turns out to be 3.35 radians. This means that the two waves are out of phase by 3.35 radians when they reach the observer.

To find the frequency closest to 300Hz to which the oscillator may be adjusted such that the observer hears minimal sound, we need to find the frequency at which the two waves will interfere destructively. This occurs when the phase difference between the two waves is (2n+1)\pi radians, where n is an integer. In this case, the closest frequency to 300Hz at which this happens is 299.7Hz.

2.) To find the points along the line joining the two speakers where relative minima of sound pressure amplitude would be expected, we can use the equation d=\frac{(2n+1)\lambda}{2}, where d is the distance from the speaker, n is an integer, and \lambda is the wavelength. In this case, the wavelength is 0.375m (calculated using \frac{v}{f}). Plugging in different values of n, we can find the distances at which destructive interference will occur. For example, when n=1, the distance is 0.563m from either speaker.

3.) To determine the maximum transverse position of an element in the medium at 0.250cm, we need to find the amplitude of the superposition of the two waves. This can be done by adding the two waves together, which results in y = 6sin\pi x cos(0.6t). The maximum amplitude occurs when cos(0.6t) is at its maximum value of 1, so the maximum transverse position would be 6cm.

4.) To approach this problem, we can use the fact that the light rays traveling from the bubble to the observer must undergo total internal reflection at the air-glass interface of the cube for the bubble to be invisible. This means that the