Basic physics/ wavelength and displacement questions

AI Thread Summary
The discussion focuses on understanding the physics of waves, specifically related to a guitar string vibrating at a fundamental frequency of 264 Hz. Participants clarify that the wavelength of the string can be calculated using the formula v = fλ, where the speed of sound in air is also considered for sound wavelength. There is confusion about the application of equations, particularly regarding the speed of light versus sound, and the concept of standing waves with nodes and antinodes is emphasized. The conversation also touches on additional questions about photon momentum and the differences between momentum and energy. Overall, the participants express a desire for deeper understanding and clarification of these physics concepts.
Noirchat
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Hi guys, i'd really like to understand physics but for some reason I'm not coping. Not sure if it's the calculations or something. But I'm willing to learn, I'd appreciate some help with this problem:

Homework Statement



A B string on a guitar with a length of 33cm is held fixed at both ends under tension. Once plucked, it ossicilates ata fundamental frequency of 264Hz.

i)What is the wavelength of the wave on the string?
ii) Sketch the pattern of displacement of the string.
iii) What is teh wavelength of the resulting sound in the air (assume the speed of sound in dry air at 0oC is 331m.s^-1)
------------
no equations were provided with the question.

-----------
my attempt:

i) c = f x lambda

lambda = c/f

lambda = 3 x 10^8/264

= 1136363.636

ii) not sure how to draw this... is there an online program so i can show my attempt?iii) lambda = 331/264

= 1.25
 
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Noirchat said:
-----------
my attempt:

i) c = f x lambda

lambda = c/f

lambda = 3 x 10^8/264
What does the speed of light have to do with this problem?

The question is asking for the wavelength of the vibrating string. The string vibrates as a transverse standing wave. A standing wave has points that do not move - nodes. The fundamental represents the lowest frequency or longest wavelength of this transverse standing wave. So the fundamental vibration occurs with the ends of the string representing consecutive nodes. How much of a wavelength is there between consecutive nodes?

ii) not sure how to draw this... is there an online program so i can show my attempt?
Can you draw a standing transverse sine wave?

AM
 
You do not need to use any equations until (iii). The equation you have used before c=λf is applied to light, this equation is derived from v=λf (seeing how with light v=c, as c is the velocity of light). v=λf can be applied to other waves (including sound). You are given the Velocity of sound in the question.

With questions (i) and (ii) i think you just need to increase your knowledge of standing waves using your book or the internet.

Hope this helps :)
 
Andrew Mason said:
What does the speed of light have to do with this problem?

The question is asking for the wavelength of the vibrating string. The string vibrates as a transverse standing wave. A standing wave has points that do not move - nodes. The fundamental represents the lowest frequency or longest wavelength of this transverse standing wave. So the fundamental vibration occurs with the ends of the string representing consecutive nodes. How much of a wavelength is there between consecutive nodes?

Can you draw a standing transverse sine wave?

AM

Saxby said:
You do not need to use any equations until (iii). The equation you have used before c=λf is applied to light, this equation is derived from v=λf (seeing how with light v=c, as c is the velocity of light). v=λf can be applied to other waves (including sound). You are given the Velocity of sound in the question.

With questions (i) and (ii) i think you just need to increase your knowledge of standing waves using your book or the internet.

Hope this helps :)
Thanks for your reply guys. I actually didn't know c=fλ only applied to electromagnetic waves! And wasn't sure what the nodes/anti-nodes did.

I have tried answering the questions again:

A B string on a guitar with a length of 33cm is held fixed at both ends under tension. Once plucked, it ossicilates ata fundamental frequency of 264Hz.

i)What is the wavelength of the wave on the string?
ii) Sketch the pattern of displacement of the string.
iii) What is teh wavelength of the resulting sound in the air (assume the speed of sound in dry air at 0oC is 331m.s^-1)
i) v=fλ , the question has given me 33cm which I've converted to 0.33 and it's given me 264 Hz.

λ = 0.33/264
=1/800 or 1.25 x 10^-3

ii) I assume this will look like a normal sine wave with the line starting at '0'.

iii) For this, is it right to say that wavelength changes with temperature since the Speed of Sound also changes with temperature? (And that pressure has no effect on the wavelength?)
An additional question:

An FM radio station broadcasts at a frequency of 109 MHz. The power radiated from the antenna is 39 kW.
a) What is the value of the momentum of each radiated photon?
b) How fast would an electron be traveling to have the same momentum? Comment on this result.

^ For this question, whent hey say 'momentum' does it mean the same as energy?
 
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Noirchat said:
i) v=fλ , the question has given me 33cm which I've converted to 0.33 and it's given me 264 Hz.

λ = 0.33/264
=1/800 or 1.25 x 10^-3
??! This makes no sense. 33 cm is a length not a speed. You seem to be ignoring my last post.

ii) I assume this will look like a normal sine wave with the line starting at '0'.
The question is: How much of a "normal sine wave" wavelength does the string represent?
iii) For this, is it right to say that wavelength changes with temperature since the Speed of Sound also changes with temperature? (And that pressure has no effect on the wavelength?)
First of all, what does this have to do with the question? (You already answered this question correctly). The speed of sound in air changes with temperature. That is true. But pressure does have a small effect because air is not quite an ideal gas.

An additional question:

An FM radio station broadcasts at a frequency of 109 MHz. The power radiated from the antenna is 39 kW.
a) What is the value of the momentum of each radiated photon?
b) How fast would an electron be traveling to have the same momentum? Comment on this result.

^ For this question, whent hey say 'momentum' does it mean the same as energy?
This seems to be a different area. What are the applicable equations? If you do not understand the difference yet between momentum and energy, your teacher should not be giving you this kind of homework!

AM
 
Like i said before, you do not need to do any calculations until (iii).

I have found a video that seems to cover pretty much everything you need to know for parts (i) and (ii) of this question.

For (iii) make sure you use the equation v=fλ, good luck.
 
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Andrew Mason said:
??! This makes no sense. 33 cm is a length not a speed. You seem to be ignoring my last post.

The question is: How much of a "normal sine wave" wavelength does the string represent?
First of all, what does this have to do with the question? (You already answered this question correctly). The speed of sound in air changes with temperature. That is true. But pressure does have a small effect because air is not quite an ideal gas.

This seems to be a different area. What are the applicable equations? If you do not understand the difference yet between momentum and energy, your teacher should not be giving you this kind of homework!

AM

I was thinking about m/s but nevermind that!

Does it represent half a 'normal sine wave'?

I'm actually learning quite a bit, so sorry if this drives you a bit crazy... is it right to say for part (i) that half a wave = 33cm. And therefore a full wave = (which equals one wavelength) = 66cm?

Thanks for clarifying it about the pressure, just wanted to make sure the info there was right.


There were no equations provided. Guess that's what one gets when they have a teacher that can't teach. I'm all on my own really.

Saxby said:
Like i said before, you do not need to do any calculations until (iii).

I have found a video that seems to cover pretty much everything you need to know for parts (i) and (ii) of this question.

For (iii) make sure you use the equation v=fλ, good luck.



Thanks for the vid, it cleared up a lot of things!

So (iii) is simply;

v=fλ

Where v = 331
f= 264

so λ = 1.25
 
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Noirchat said:
I was thinking about m/s but nevermind that!

Does it represent half a 'normal sine wave'?

I'm actually learning quite a bit, so sorry if this drives you a bit crazy... is it right to say for part (i) that half a wave = 33cm. And therefore a full wave = (which equals one wavelength) = 66cm?
Very good.

There were no equations provided.
The momentum of a photon is h/λ or hv/c. An electron's momentum is mv where m is the mass of an electron, which you can look up and v is its velocity. You should be able to work out v from that.
Guess that's what one gets when they have a teacher that can't teach. I'm all on my own really.
Good teachers encourage you think. The thinking, though, you have to do on your own. Good luck.

AM
 
Noirchat said:
I'm actually learning quite a bit, so sorry if this drives you a bit crazy... is it right to say for part (i) that half a wave = 33cm. And therefore a full wave = (which equals one wavelength) = 66cm?

So (iii) is simply;

v=fλ

Where v = 331
f= 264

so λ = 1.25

Well done, You've got both parts (i) and (iii) correct. I assume that you know how to draw the diagram in part (ii)?
 
  • #10
Andrew Mason said:
Very good.

The momentum of a photon is h/λ or hv/c. An electron's momentum is mv where m is the mass of an electron, which you can look up and v is its velocity. You should be able to work out v from that.
Good teachers encourage you think. The thinking, though, you have to do on your own. Good luck.

AM

Thank you! :)

So for part (a),

λ=v/f

λ = 3 x 10^8/10490000
λ = 28.599

Momentum of photon = 6.626 x 10^-34/28.599

=2.32 x 10^-35

(b)
momentum of electron = mv

v = p/m

where p = 2.32 x 10^-35
m = 9.10938 x 10^-31

v = 2.5468 x 10^-5

That is very true, i don't want answers straight out, i prefer explanations and reasoning behind it. It'll definitely help if i come across such questions again.
Thanks a lot Andrew. :)

Saxby said:
Well done, You've got both parts (i) and (iii) correct. I assume that you know how to draw the diagram in part (ii)?

For part (ii) I think it's the first half of a y=sinx graph?
 
  • #11
Noirchat said:
Thank you! :)

So for part (a),

λ=v/f

λ = 3 x 10^8/10490000
λ = 28.599

Momentum of photon = 6.626 x 10^-34/28.599

=2.32 x 10^-35
You have the right approach but you said the frequency is 1.09x10^8 s^-1 not 1.049x10^7

Also, be sure to state the units of your answer (metres).

(b)
momentum of electron = mv

v = p/m

where p = 2.32 x 10^-35
m = 9.10938 x 10^-31

v = 2.5468 x 10^-5
Right approach. When you work out the correct momentum you can get the correct answer.

For part (ii) I think it's the first half of a y=sinx graph?
Almost correct - right idea. What is the maximum amplitude or displacement of the string from its flat position? How does that factor into your equation?

AM
 
  • #12
Andrew Mason said:
You have the right approach but you said the frequency is 1.09x10^8 s^-1 not 1.049x10^7

Also, be sure to state the units of your answer (metres).

Right approach. When you work out the correct momentum you can get the correct answer.

Almost correct - right idea. What is the maximum amplitude or displacement of the string from its flat position? How does that factor into your equation?

AM

Got the frequency from another q mixed up.

Again; (a)

λ=v/f

λ = 3 x 10^8/109000000 = 2.7523

Momentum of photon = 6.626 x 10^-34/2.7523

=2.407 x 10 ^-34

(b)
momentum of electron = mv

v = p/m

p =2.407 x 10 ^-34
m = 9.10938 x 10^-31

v = 2.642 x 10^-66 m/s

///


It would be at the anti-node?
 
  • #13
Noirchat said:
Got the frequency from another q mixed up.

Again; (a)

λ=v/f

λ = 3 x 10^8/109000000 = 2.7523

Momentum of photon = 6.626 x 10^-34/2.7523

=2.407 x 10 ^-34

(b)
momentum of electron = mv

v = p/m

p =2.407 x 10 ^-34
m = 9.10938 x 10^-31

v = 2.642 x 10^-66 m/s

///


It would be at the anti-node?
Yes.

The problem is that the maximum value of y in y = sin x is 1. Let's say that the maximum amplitude is A (ie. this is the maximum value of y). How would you change the equation y = sin x so that y = A at the peak of that sine curve?

AM
 
  • #14
Andrew Mason said:
Yes.

The problem is that the maximum value of y in y = sin x is 1. Let's say that the maximum amplitude is A (ie. this is the maximum value of y). How would you change the equation y = sin x so that y = A at the peak of that sine curve?

AM

Well since the numeral/letter in front of 'sin' gives you the range of the graph wouldn't it be: y = Asinx?
 
  • #15
Noirchat said:
Well since the numeral/letter in front of 'sin' gives you the range of the graph wouldn't it be: y = Asinx?
Very good. That will give you the equation for the shape of the string at an instant in time where the mid point of the string is has a displacement A from zero.

AM
 
  • #16
Andrew Mason said:
Very good. That will give you the equation for the shape of the string at an instant in time where the mid point of the string is has a displacement A from zero.

AM

I assume the graph will look like this:
http://media.smashingmagazine.com/wp-content/uploads/2011/09/sine.png. ?
 
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