Query Asst 25

course Phy 122

o?????????h??????assignment #025025. `Query 14

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

Physics II

07-11-2007

......!!!!!!!!...................................

00:20:17

Query Principles of Physics and General College Physics 12.40: Beat frequency at 262 and 277 Hz; beat frequency two octaves lower.

......!!!!!!!!...................................

RESPONSE -->

The beat frequency will be the difference in the frequencies of the two pitches:

f_beat = 277Hz - 262Hz = 15Hz

Therefore, the beat frequency is 15 beats/sec.

If each pitch is played two octaves lower, the frequencies will be reduced by a factor of 4. In order to find the new beat frequency, we simply divide 15 beats/sec by 4:

15Hz / 4 = 3.75Hz

.................................................

......!!!!!!!!...................................

00:20:31

The beat frequency is the difference in the frequencies, in this case 277 Hz - 262 Hz = 15 Hz.

One ocatave reduces frequency by half, so two octaves lower would give frequencies 1/4 as great. The difference in the frequencies would therefore also be 1/4 as great, resulting in a beat frequency of 1/4 * 15 Hz = 3.75 Hz.

......!!!!!!!!...................................

RESPONSE -->

I understand this problem and I answered it correctly.

.................................................

......!!!!!!!!...................................

00:20:49

Query gen phy problem 12.46 speakers 1.8 meters apart, listener three meters from one and 3.5 m from the other **** gen phy what is the lowest frequency that will permit destructive interference at the location of the listener?

......!!!!!!!!...................................

RESPONSE -->

I am not a general physics student and this problem was not assigned to me.

.................................................

......!!!!!!!!...................................

00:20:59

STUDENT SOLUTION: To solve this problem, I first realize that for destructive interference to occur, the path difference is an odd multiple of half of the wavelength(ex. 1/2, 3/2, 5/2). I used Fig.12-17 in the text to help visualize this problem. Part (A) of the problem asks to calculate the lowest frequency at which destructive interference will occur at the point where two loudspeakers are 2.5m apart and one person stands 3.0m from one speaker and 3.5m from the other. The text states that 'destructive interference occurs at any point whose distance from one speaker is greater than its distance from the other speaker by exactly one-half wavelength.' The path difference in the problem is fixed, therefore the lowest frequency at which destructive interference will occur is directly related to the longest wavelength. To calculate the lowest frequency, I first have to calculate the longest wavelength using the equation

'dL ='lambda/2, where `dL is the path difference.

'lambda=2*'dL

=2(3.5m-3.0m)=1m

Now I can calculate the frequency using

f=v/'lambda. The velocity is 343m/s which is the speed of sound.

f=343m/s/1m=343 Hz.

Thus, the lowest frequency at which destructive interference can occur is at 343Hz.

Keeping in mind that destructive interference occurs if the distance equals an odd multiple of the wavelength, I can calculate (B) part of the problem.

To determine the next wavelength, I use the equation 'dL=3'lambda/2

wavelength=2/3(3.5m-3.0m) =0.33m

Now I calculate the next highest frequency using the equation f=v/wavelength.

f^2=343m/s/0.33m=1030Hz.

I finally calculate the next highest frequency.

'del L=5/2 'lambda

wavelength=0.20m

f^3=343m/s/0.2m=1715 Hz.

INSTRUCTOR EXPLANATION:

The listener is .5 meters further from one speaker than from the other. If this .5 meter difference results in a half-wavelength lag in the sound from the further speaker, the peaks from the first speaker will meet the troughs from the second. If a half-wavelength corresponds to .5 meters, then the wavelength must be 1 meter. The frequency of a sound with a 1-meter wavelength moving at 343 m/s will be 343 cycles/sec, or 343 Hz.

The next two wavelengths that would result in destructive interference would have 1.5 and 2.5 wavelengths corresponding to the .5 m path difference. The wavelengths would therefore be .5 m / (1.5) = .33 m and .5 m / (2.5) = .2 m, with corresponding frequencies 343 m/s / (.33 m) = 1030 Hz and 343 m/s / (.2 m) = 1720 Hz, approx. ****

......!!!!!!!!...................................

RESPONSE -->

Ok.

.................................................

......!!!!!!!!...................................

00:22:07

**** gen phy why is there no highest frequency that will permit destructive interference?

......!!!!!!!!...................................

RESPONSE -->

I am not a general physics student, but I would like to attempt to answer this question.

Destructive interference can occur at any frequency, as long as the wavelengths are a number and a half.

.................................................

......!!!!!!!!...................................

00:23:08

** You can get any number of half-wavelengths into that .5 meter path difference. **

......!!!!!!!!...................................

RESPONSE -->

I don't really understand this solution. What is the .5m path difference? Maybe this was a follow-up question, and in that case I wouldn't have been able to answer it exactly right. I feel that I understand the concept of destructive interference, though.

One speaker is .5 meters further from the listener than the other.

Destructive interference occurs when one wave follows a path which is 1/2, or 3/2, or 5/2, etc. wavelength (in general a whole number of wavelengths plus half of a wavelength) greater than the other.

For example a wavelength of .1 meter would have 5 wavelengths in the .5 meter distance and would constructively interfere; to get 5.5 wavelengths and destructive interference the wavelength would be .5 m / 5.5 = .091 meter, approx.

If we pick any whole number n, we get constructive interference for wavelength .5 m / n and destructive interference for wavelength .5 m / (n + 1/2).

.................................................

......!!!!!!!!...................................

00:24:17

gen phy what must happen in order for the sounds from the two speakers to interfere destructively, assuming that the sources are in phase?

......!!!!!!!!...................................

RESPONSE -->

I'm not general physics, but I feel like I can answer this question:

Chapter 12 says that if the distance between the speakers is equal to precisely 1/2 the wavelength of the sound, the waves will be out of phase when they meet and destructive interference will occur.

Good, but the key is the difference between the distances. If one speaker is 1/2 wavelength further from the listener than the other, destructive interference will result.

.................................................

......!!!!!!!!...................................

00:24:29

** The path difference has to be and integer number of wavelengths plus a half wavelength. **

......!!!!!!!!...................................

RESPONSE -->

I feel that I understand this concept.

.................................................

......!!!!!!!!...................................

00:24:55

CRAB NEBULA PROBLEM?

......!!!!!!!!...................................

RESPONSE -->

I am not familiar with the crab nebula problem and cannot answer this question.

.................................................

......!!!!!!!!...................................

00:25:01

This Query will exit.

......!!!!!!!!...................................

RESPONSE -->

.................................................

"

&#

Good responses. See my notes and let me know if you have questions. &#