Interference of waves (check my hmwk questions please)

In summary, the conversation revolved around solving a question involving frequency and superposition. The participants discussed various equations and methods to find the correct answers, including using rubber bands to increase the mass and slow down the frequency of a tuning fork. They also shared their attempts and explanations for the superposition question and provided suggestions for improvement.
  • #1
supernova1203
210
0
For the first question, number 8 i have put up 2 attachment, one with the question the other one with the answer, I am not sure of the answers however, so if someone can check to see if theyre right or not, and if not show me how to attain the right answer.

thanks
For the next question, number 11, ill just show you the question and do the solution right here.

11)a) f1= 320 Hz
f2=?

f=# of cycles/total time

f=14/7

f= 2Hz (this 2 hz is also the beat frequency)

fb is beat frequency

fb= f2-f1 there's a vertical line for absolute value but i don't know how to type it in

f2= f1-fb

f2=320 + or - 2

f2= 322 hz or f= 318 hzthat was 11a

now 11b11b) f1=320 hz
fb=1 hz(the beat frequency if you looked at question is now 1 instead of previous 2 hz)

fb = f2-f1

f2=f1-fb

f2= 320 + or - 1

f2= 319 hz or f2=321 hzNow how do i know which is the original frequency of tuning fork 2? i don't know what to do next, i don't know which is the right frequency for tuning fork 2.
 

Attachments

  • Lesson 3 question 1.jpg
    Lesson 3 question 1.jpg
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  • Check 1.jpg
    Check 1.jpg
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  • Lesson 3 question 2.jpg
    Lesson 3 question 2.jpg
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Last edited:
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  • #2


anyone?
 
  • #3


supernova1203 said:
For the first question, number 8 i have put up 2 attachment, one with the question the other one with the answer, I am not sure of the answers however, so if someone can check to see if theyre right or not, and if not show me how to attain the right answer.

thanks



For the next question, number 11, ill just show you the question and do the solution right here.

11)a) f1= 320 Hz
f2=?

f=# of cycles/total time

f=14/7

f= 2Hz (this 2 hz is also the beat frequency)

fb is beat frequency

fb= f2-f1 there's a vertical line for absolute value but i don't know how to type it in

f2= f1-fb

f2=320 + or - 2

f2= 322 hz or f= 318 hz


that was 11a

now 11b


11b) f1=320 hz
fb=1 hz(the beat frequency if you looked at question is now 1 instead of previous 2 hz)

fb = f2-f1

f2=f1-fb

f2= 320 + or - 1

f2= 319 hz or f2=321 hz


Now how do i know which is the original frequency of tuning fork 2? i don't know what to do next, i don't know which is the right frequency for tuning fork 2.

Adding the rubber bands increases the mass of the fork which slows it down, so the second frequency is smaller than the first.
 
  • #4


supernova1203 said:
anyone?

I didn't like the look of you superposition.

It appears that one pule is trying to take the string down, while the other is trying to take the string up. Net result can tend to be very little.

You are adding these pulses [looks like you may have subtracted].

Given that the "down" pulse is a square pulse, you can simply translate the "up" pulse lower.

I suspect a slight up result at the left, and then a mainly down result at the RHS.
 
  • #5


hm.. ok i tried the super positioning question again, let me know what you think, and if i don't have it right can you draw it on paint and post it here so i can see what you did, and preferably how you did it?
(new attempt in new attachment)

this is the new attempt at question 8, the very first attachment from original question.an explanation is in order i suppose, the superposition looks the way it does because negative and positive values are assigned to top and bottom, since there is more at bottom half, we assume there is a higher negative value attached to it, and so i basically shifted it into the negative area(which is below the line or rest position)
 

Attachments

  • Best attempt super positioning.jpg
    Best attempt super positioning.jpg
    1.7 KB · Views: 475
  • #6


supernova1203 said:
hm.. ok i tried the super positioning question again, let me know what you think, and if i don't have it right can you draw it on paint and post it here so i can see what you did, and preferably how you did it?



(new attempt in new attachment)

this is the new attempt at question 8, the very first attachment from original question.


an explanation is in order i suppose, the superposition looks the way it does because negative and positive values are assigned to top and bottom, since there is more at bottom half, we assume there is a higher negative value attached to it, and so i basically shifted it into the negative area(which is below the line or rest position)

I agree with the principle of your answer, but when I look at the original problem, I think the upward/sloping pulse has a maximum which is larger than the downward/square pulse, so I think the sloping resultant you are drawing should start out a little upward and slope down the way you show.

If I were to put values on the pulses - for explanation purposes - I would say the up ward pulse has a peak value of 3, sloping down to 0, while the downward pulse is a steady -2. The resultant would start at a peak value of 1, sloping down to -2.

Perhaps those values are a little extreme [in comparison] and It should have been 5 → 0 for the upper pulse, and -4 for the lower, with the resultant going from 1 → -4.

Those values are arbitrary, just used to show relative size of the pulses.
 
  • #7


PeterO said:
I agree with the principle of your answer, but when I look at the original problem, I think the upward/sloping pulse has a maximum which is larger than the downward/square pulse, so I think the sloping resultant you are drawing should start out a little upward and slope down the way you show.

If I were to put values on the pulses - for explanation purposes - I would say the up ward pulse has a peak value of 3, sloping down to 0, while the downward pulse is a steady -2. The resultant would start at a peak value of 1, sloping down to -2.

Perhaps those values are a little extreme [in comparison] and It should have been 5 → 0 for the upper pulse, and -4 for the lower, with the resultant going from 1 → -4.

Those values are arbitrary, just used to show relative size of the pulses.

the problem is that in the course so far they haven't assigned any numerical or mathematical values to any of these waves or their superpositions, they haven't even mentioned them, they just expect me to somehow solve them visually just be looking at them, and without numbers or mathematics that's the best i can come up with, however if they did use mathematics/numbers i would have been able to solve them much more accurately and easily.
 
  • #8


supernova1203 said:
the problem is that in the course so far they haven't assigned any numerical or mathematical values to any of these waves or their superpositions, they haven't even mentioned them, they just expect me to somehow solve them visually just be looking at them, and without numbers or mathematics that's the best i can come up with, however if they did use mathematics/numbers i would have been able to solve them much more accurately and easily.

That is OK, but I used the numbers - because I can't draw on this forum - as a way of describing what I thought the diagram looked like. The higher end of the upward pulse looked bigger than how far the lower pulse goes down, so the resultant should have a small upward part to the pulse at the left end.

If you have the diagram on paper, you could actually use a ruler and measure the size. If only electronic, you could copy it into "paint" or similar and drag the image around and compare it for size.
 
  • #9


PeterO said:
That is OK, but I used the numbers - because I can't draw on this forum - as a way of describing what I thought the diagram looked like. The higher end of the upward pulse looked bigger than how far the lower pulse goes down, so the resultant should have a small upward part to the pulse at the left end.

If you have the diagram on paper, you could actually use a ruler and measure the size. If only electronic, you could copy it into "paint" or similar and drag the image around and compare it for size.

I did the "paint" thing and copied the left hand end of the lower pulse into the middle of the upper pulse, confirming that the peak of the upper pulse is bigger than the lower pulse.
If I understand how this works, the attachment should be here as well.

EDIT: That worked well. You could actually do the supoerposition in paint, by copying the upper pulse and pasting it onto the lower pulse in a similar way.
 

Attachments

  • Superposition.jpg
    Superposition.jpg
    7.4 KB · Views: 453

Related to Interference of waves (check my hmwk questions please)

1. What is interference of waves?

Interference of waves is a phenomenon that occurs when two or more waves meet and interact with each other. It can result in either constructive interference, where the waves combine to form a larger amplitude, or destructive interference, where the waves cancel each other out.

2. What are some real-life examples of interference of waves?

Some examples of interference of waves in everyday life include sound waves from two speakers overlapping and creating a louder sound, light waves from a soap bubble creating a colorful pattern due to interference, and ocean waves interacting with each other to create larger or smaller waves.

3. How does the wavelength of a wave affect interference?

The wavelength of a wave determines the distance between the peaks and troughs of a wave. When two waves with the same wavelength meet, they can create either constructive or destructive interference. If the waves have different wavelengths, the interference will be more complex and can result in a combination of constructive and destructive interference.

4. Can interference of waves be controlled or manipulated?

Yes, interference of waves can be controlled and manipulated in various ways. This is the basis for technologies such as noise-canceling headphones, which use interference to cancel out unwanted sound waves. It is also used in scientific experiments to study the properties of waves and in engineering to improve the performance of devices such as antennas and lasers.

5. How does the angle of incidence affect interference of waves?

The angle of incidence, or the angle at which waves meet, can affect the interference pattern that is created. When waves meet at an angle, the resulting interference can create a pattern of alternating bright and dark fringes. This is known as diffraction and is commonly observed in experiments with light waves passing through narrow slits or small objects.

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