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Homework Help: Sound Interference + Light

  1. Oct 17, 2005 #1
    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. [itex]\Delta r = \frac{\phi}{2\pi}\lambda[/itex] and [itex]\lambda = \frac{v}{f}[/itex], hence i solved for [itex]\phi[/itex], 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 [itex] (2n+1)\pi[/itex]. 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 [itex]\frac{v}{f}[/itex]. Thus wavelength equals 2.33m. Then i used the equation [itex]d = \frac{(2n+1)\lambda}{2}[/itex]. Is this correct? What do i solve for? Where do i go from there?

    3.) Two sinusoidal waves combine in medium are described by:
    [itex]y = 3sin\pi(x + 0.6t)[/tex]
    [itex]y = 3sin\pi(x - 0.6t)[/tex]
    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 im 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 couldnt get a clear understanding of the situation. If anyone could draw me a sample diagram, i would be very grateful.
  2. jcsd
  3. Oct 22, 2005 #2
    No one?

  4. Oct 23, 2005 #3
    I'm unsure as to why no one can help me on this.
  5. Oct 25, 2005 #4
    If anyone can offer any kind of help, it would be greatly appreciated.

    You don't have to answer all the questions at once, but even an answer to one of them will help me greatly.

    Thank you.
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