Maximizing Profit in Online Business

In summary: L/3.In summary, the author suggests that if you touch the string lightly at a position L/3, the string will continue to oscillate as a whole.
  • #1
maxelcat
33
4
Homework Statement
The fundamental frequency f is the lowest frequency heard when a stretched string is vibrating.
The string is now lightly touched one third of the way along its length.
What is the lowest frequency heard?

The answer seems to be 3f - but I cannot get that.
Relevant Equations
speed = freq x wavelength
Here's my working. Perhaps I make some wrong assumptions?
1618331589559.png
 
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  • #2
Your reasoning is not clear to me. Also, what does "lightly touched" mean? Is it supposed to imply that the string needs to have a node there? (Note that the wave with wavelength 4L/3 does not have a node at one of the ends.
 
  • #3
well, I read it as being a node at the point at the point where it is touched
 
  • #4
Orodruin said:
Also, what does "lightly touched" mean?

I believe the "lightly" qualifier for these types of problems serves to indicate that the parts of the string on both sides of the node continue to oscillate,

as opposed to having the string on one particular side of the node stop oscillating, effecting a reduction in the oscillating length.

(e.g. as shown here: https://commons.wikimedia.org/wiki/File:Flageolette.svg)
 
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  • #5
maxelcat said:
well, I read it as being a node at the point at the point where it is touched
Is there any frequency with wavelength 4L/3 for the string of length L?
 
  • #6
etotheipi said:
I believe the "lightly" qualifier for these types of problems serves to indicate that the parts of the string on both sides of the node continue to oscillate,

as opposed to having the string on one particular side of the node stop oscillating, effecting a reduction in the oscillating length.

(e.g. as shown here: https://commons.wikimedia.org/wiki/File:Flageolette.svg)
that is how i read it. so string vibrates in two sections one of lenth 2L/3 and one L/3
 
  • #7
maxelcat said:
that is how i read it. so string vibrates in two sections one of lenth 2L/3 and one L/3
No, that is not what he said. That would be holding the string firmly at L/3.
 
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  • #8
From my understanding of the problem, ##f_2## is your frequency for the string of length ##2/3L##. Try to isolate it on one side of the equal sign and see what frequency you get.

Then do the problem for the string of length ##1/3L## and see what frequency you get for that.
 
  • #9
Orodruin said:
No, that is not what he said. That would be holding the string firmly at L/3.
so that would mean there is only going to wave - the shorter one
 
  • #10
hicetnunc said:
From my understanding of the problem, ##f_2## is your frequency for the string of length ##2/3L##. Try to isolate it on one side of the equal sign and see what frequency you get.

Then do the problem for the string of length ##1/3L## and see what frequency you get for that.
Again, this is not how I read the problem based on #4.
 
  • #11
Orodruin said:
Again, this is not how I read the problem based on #4.
I think the problem has been WAY overanalysed. I think it's just a problem involving an initial string of length ##L## and frequency ##f##, that you subdivide into two parts and calculate their frequencies. At least, that would be my experience with homework problems.

For any deeper analytics, one would need more information about the string.
 
  • #12
@maxelcat, the point is that if you touch the string lightly at a position ##L/3##, then the string continues to oscillate "as a whole", i.e. the new lowest frequency of oscillation (say ##f'##) is the same for all points on the string. You are essentially just imposing the condition that the string, of length ##L##, has a node at position ##L/3##. Have a look at case 5 in the image I linked to in post #4; it should then be clear why ##f' = 3f##!

This is not the same as pressing firmly, which would result in the two sides of the string becoming independent of each other (effectively, you'd instead have two separate, shorter strings).
 
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  • #13
etotheipi said:
@maxelcat, the point is that if you touch the string lightly at a position ##L/3##, then the string continues to oscillate "as a whole", i.e. the new lowest frequency of oscillation (say ##f'##) is the same for all points on the string. You are essentially just imposing the condition that the string, of length ##L##, has a node at position ##L/3##. Have a look at case 5 in the image I linked to in post #4; it should then be clear why ##f' = 3f##!

This is not the same as pressing firmly, which would result in the two sides of the string becoming independent of each other (effectively, you'd instead have two separate, shorter strings).
Yes - I can see that. I misunderstood what it meant about 'lightly'. Thanks all. (BTW this is an AS level q so it shouldn't be hard!)
 
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  • #14
hicetnunc said:
I think the problem has been WAY overanalysed. I think it's just a problem involving an initial string of length ##L## and frequency ##f##, that you subdivide into two parts and calculate their frequencies. At least, that would be my experience with homework problems.

For any deeper analytics, one would need more information about the string.
No, you just have the wrong interpretation. The entire idea of being touched lightly is to force the string, as a whole, into an oscillation that has a node point at L/3.
 

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