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Physics Problem -- PLEASE HELP!!!!!

 
Old 04-22-2011 at 10:06 AM   #16
Faer
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Q. A student uses an audio oscillator of adjustable frequency to measure the depth of a dried up well. Two successive resonances are heard at 28.5 Hz and 39.9 Hz. The speed of sound in air is 343 m/s. How deep is the well?

So I solved this by equating f1 and f2 with m(v/4L), with m= m and (m+1) respectively. The problem is, I get L as exactly half of the answer I'm supposed to get! I'm pretty sure that the formula is NOT supposed to be v/2L, since this is an open-closed tube. Could anyone tell me where I went wrong?
Old 04-22-2011 at 10:08 AM   #17
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Quote:
Originally Posted by Faer View Post
Q. A student uses an audio oscillator of adjustable frequency to measure the depth of a dried up well. Two successive resonances are heard at 28.5 Hz and 39.9 Hz. The speed of sound in air is 343 m/s. How deep is the well?

So I solved this by equating f1 and f2 with m(v/4L), with m= m and (m+1) respectively. The problem is, I get L as exactly half of the answer I'm supposed to get! I'm pretty sure that the formula is NOT supposed to be v/2L, since this is an open-closed tube. Could anyone tell me where I went wrong?
It's actually a close-closed tube. The oscillator you're holding is a closed end.

I had the same problem. =P

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Old 04-22-2011 at 10:10 AM   #18
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NO FREAKIN' WAY. That's ABSURD! OMG.

Shooooooot...LOL. Thank you so much! What an incredible oversight... >.>

EDIT:

Waves from a radio station have a wavelength of 326 m. They travel by two paths to a home receiver 20.6 km from the transmitter. One path is a direct path, and the other is by reflection from a mountain directly behind the home receiver. What is the minimum distance from the mountain to the receiver that produces destructive interference at the receiver? (Assume that no phase change occurs on reflection from the mountain.)


Last edited by Faer : 04-22-2011 at 10:42 AM.
Old 04-22-2011 at 11:09 AM   #19
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Quote:
Originally Posted by Faer View Post
NO FREAKIN' WAY. That's ABSURD! OMG.

Shooooooot...LOL. Thank you so much! What an incredible oversight... >.>

EDIT:

Waves from a radio station have a wavelength of 326 m. They travel by two paths to a home receiver 20.6 km from the transmitter. One path is a direct path, and the other is by reflection from a mountain directly behind the home receiver. What is the minimum distance from the mountain to the receiver that produces destructive interference at the receiver? (Assume that no phase change occurs on reflection from the mountain.)

Ok, NORMALLY, you would have to find the phase of the first wave, the phase of the second wave, the resulting phase difference, and since we want destructive interference, equate the phase difference to (m+1/2)2pi, where m=0,1,2,3.... In this case, m=0, since we want the minimum distance. Then solve for the path difference, and divide the path difference by 2 (the distance from the mountain to the receiver is HALF the path difference!)

NORMALLY, you would do that. But note the assumption in parentheses: "Assume that no phase change occurs on reflection from the mountain." This means that the INITIAL PHASE DIFFERENCE is zero (remember, the phase difference = 2pi(path difference / wavelength) + initial phase difference).**

Since the initial phase difference is zero, the phase difference between the two waves is simply 2pi(path difference/wavelength). This means we can skip everything and use the destructive interference formula path difference = (m+1/2)wavelength. Since m=0 (we want the minimum path difference), path difference=wavelength /2. Then divide that path difference by 2 to get the distance from the receiver to the mountain.

**Note: If there was a phase change upon reflection from the mountain, the situation would be different. It would be as follows:

phase of wave 1 = 0
phase of wave 2 = 2pi(path difference/wavelength) + pi
phase difference = 2pi(path difference/wavelength) + pi = (m+1/2)2pi

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Old 04-22-2011 at 11:14 AM   #20
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Quote:
Originally Posted by Faer View Post
Waves from a radio station have a wavelength of 326 m. They travel by two paths to a home receiver 20.6 km from the transmitter. One path is a direct path, and the other is by reflection from a mountain directly behind the home receiver. What is the minimum distance from the mountain to the receiver that produces destructive interference at the receiver? (Assume that no phase change occurs on reflection from the mountain.)
For this question, you need to think back to what causes destructive interference and how it can be applied to this question in particular.

--> Equation for destructive interference: dx = (m + 1/2)lambda
--> Equation for constructive interference: dx = m*lambda
** The key think to note here is the 1/2 that's characteristic of destructive interference.

Now, since the transmitter emits waves that travel in all directions, you know that it will:
1.) travel directly to the home receiver
2.) travel in the opposite direction to the mountain, get reflected, and then travel back towards the receiver

In order for you to have destructive interference of those two waves, there needs to be a wavelength difference of 1/2. In order for the wave to have a wavelength difference of 1/2 by the time it reaches the other wave, the wave must travel 1/4 of a wavelength from the source to the mountain, and another 1/4 of a wavelength from the mountain to the source.

Therefore, the distance the mountain needs to be from the source is 1/4 of a wavelength.

Edit: shit, crucible beat me to it. lol
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Old 04-22-2011 at 11:45 AM   #21
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let me guess Phys 1b03 exam is 2worro or monday?
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Old 04-22-2011 at 11:58 AM   #22
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Quote:
Originally Posted by thedog123123 View Post
let me guess Phys 1b03 exam is 2worro or monday?
Tomorrow -_-

Btw, I JUST realized who you were. I've met you before, Colin! Physics meet and greet about a month ago? I was the one first year, red hair, that listened to you, Scott, Jeremie, etc. ramble on for a couple of hours.

(I'm assuming this is the same Colin, haha)

Edit... wait, I may be wrong.
Edit (2)... nevermind, I am wrong. He was in 2nd year.../talking to myself
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Last edited by tatski-p : 04-22-2011 at 12:02 PM.
Old 04-22-2011 at 12:13 PM   #23
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Could you guys help me with these two?

1. A fresh water pond that is 14.4 m deep is contained on one side by a cliff. The water has eroded a nearly horizontal "tube" through a bed of limestone, which allows the water to emerge on the other side of the cliff. If the "tube" has a diameter of 4.30 cm, and is located 6.3 m below the surface of the pond, what is the frictional force between the "tube" wall and a rock that is blocking the exit?

2.
Old 04-22-2011 at 12:22 PM   #24
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Thank you so much! That was really helpful, and just made me realized how unprepared I am...gah. -.-

Anyhow. I'll be pouring in with questions throughout today for the next...howsoever many hours it is till the exam. XD

1)A certain crude oil has an index of refraction of 1.26. A ship dumps 1.17 m3 of this oil into the ocean, and the oil spreads into a thin uniform slick. If the film produces a first-order maximum of light of wavelength 509 nm normally incident on it, how much surface area of the ocean does the oil slick cover? Assume that the index of refraction of the ocean water is 1.35.

I can't figure out the relation between n and thickness. :/

EDIT: Might as well...

2) Determine the minimum thickness of a soap film (n = 1.350) that will result in constructive interference of the red Hα line (λ=656.3 nm);

Last edited by Faer : 04-22-2011 at 01:25 PM.
Old 04-22-2011 at 12:32 PM   #25
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Quote:
Originally Posted by BVictor View Post
Could you guys help me with these two?

1. A fresh water pond that is 14.4 m deep is contained on one side by a cliff. The water has eroded a nearly horizontal "tube" through a bed of limestone, which allows the water to emerge on the other side of the cliff. If the "tube" has a diameter of 4.30 cm, and is located 6.3 m below the surface of the pond, what is the frictional force between the "tube" wall and a rock that is blocking the exit?

2.
Oooh, I know, I've done these!

1) Basically, you know F = P/a, right? You need to find the pressure difference on either side of the rock, and then divide it by the area of the tube.

Using the horizontal-line-same-pressure model, we can see that on the left side of the rock - the side fo the tube towards the pond - the pressure = p0 + (rho)gh, the pressure due to the depth of the water. On the right side of the rock, where the water is exiting from the cliff, the pressure is atmospheric pressure. Hence, you find the difference and divide by A.

For the next part of that question, you'll have to use Bernoulli's equation - but carefully. You need to apply a ton of assumptions and stuff, which I'll mention if you still need to solve it.

(2) What the question is asking is, the phase difference of the two waves in terms of no. of waves. That is, the same waves are going in, getting changes, coming out again, and you want to see how the wavelength differs and how that affects no. of waves. The first thing you need to find is the no. of wavelengths produced INSIDE the material. For n2, since it is shorter than n1, you also need to find the no. of wavelengths in the small space where the wave is in the air (the length n2 is shorter than n1). That's because you need to measure the wavelength difference at the EXACT SAME POINT (which we consider the end of n1).

You know that the wavelengths of the two waves change inside the medium, but note that the frequencies always remain the same. So the first step is to find the frequencies. Since these waves are light waves, v=3*10^8 m/s.

After you've found the frequencies, you need to find wavelengths INSIDE the medium. You use the same formula of v/f, however, note that inside the medium, v changes. You can find v using c/n, where both c and n are known.

Now that you have the wavelengths inside each respective medium, you can divide the length of the medium to get the no. of waves inside each. For n2, you have to find the no. of waves for the section inside the air too, using the initial wavelength given.

The phase difference would then be the (no. of waves from n2) - (no. of waves from n1).

Hope this helps!

Last edited by Faer : 04-22-2011 at 12:34 PM.
Old 04-22-2011 at 12:34 PM   #26
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for this one, it confuses me that the pipe is narrower at Y than X

the answer is E

Last edited by Kayee : 04-22-2011 at 01:49 PM.
Old 04-22-2011 at 01:24 PM   #27
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Fcuk physics. It doesnt help you get bitches.
Old 04-22-2011 at 02:23 PM   #28
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Struggling with this uncertainty question, someone help please :$

You drop a ball (with zero initial velocity) from the top of a building. If you measure the acceleration to be 9.8 ± 0.1 m/s2, and the height of the building to be 30±1 m, what is the final velocity of the ball?
Old 04-22-2011 at 02:44 PM   #29
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Quote:
Originally Posted by ShesTheMan View Post
Struggling with this uncertainty question, someone help please :$

You drop a ball (with zero initial velocity) from the top of a building. If you measure the acceleration to be 9.8 ± 0.1 m/s2, and the height of the building to be 30±1 m, what is the final velocity of the ball?
Is this a multiple choice question? Do you have the options? I think I may have an answer but would like to make sure before I explain it.
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Old 04-22-2011 at 02:50 PM   #30
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Quote:
Originally Posted by tatski-p View Post
Is this a multiple choice question? Do you have the options? I think I may have an answer but would like to make sure before I explain it.

Yes it is, the options are:
a) 24.2 ± 0.5 m/s
b) 24.25 ± 0.04 m/s
c) 24±1m/s
d) 24.25 ± 0.53 m/s
e) 24.2 ± 0.02 m/s



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