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Difficult but interesting standing waves question

  • Thread starter nil1996
  • Start date
  • #1
301
7

Homework Statement


A 160 g rope 4m long is fixed at one end and tied to a light string of same length to other end.Its tension is 400N.
What is the wave length of first overtone?

Homework Equations


λ=v/f
v=√(T/μ)
μ=mass per unit length
T=tension
f=frequency

The Attempt at a Solution


Well,
i know to find the wave length when the string is uniform density.Here i am unable to think how this works when we have two strings of different μ.Please guide me.First overtone is
fundamental_3.JPG
 

Answers and Replies

  • #2
199
15
Well,
i know to find the wave length when the string is uniform density.Here i am unable to think how this works when we have two strings of different μ.Please guide me.
There are some conditions for these kind of problems, known as boundary conditions.

1. Tension would be same in both the strings.
2. Frequency would be same for both strings.

when a heavy string is connected to a lighter string, that junction point is considered to be an antinode. I have given you pretty much every information you need to solve this question!!!
 
  • #3
301
7
There are some conditions for these kind of problems, known as boundary conditions.

1. Tension would be same in both the strings.
2. Frequency would be same for both strings.

when a heavy string is connected to a lighter string, that junction point is considered to be an antinode. I have given you pretty much every information you need to solve this question!!!
Nihal are you from india.I am from india.
 
  • #4
199
15
Nihal are you from india.I am from india.
Well, yes I am. But that is irrelevant to the question.:wink:
 
  • #5
301
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Well, yes I am. But that is irrelevant to the question.:wink:
are you preparing for iitjee??
 
  • #6
199
15
  • #7
301
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Why? Are you?
Yes i am preparing for iitjee without coaching.
 
  • #8
199
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Yes i am preparing for iitjee without coaching.
good luck. If you want talk about iitjee and all that, use private message.
 
  • #9
301
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Still not getting the problem
 
  • #10
199
15
Still not getting the problem
What is it that you are not getting???
 
  • #11
301
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I am not understanding how will the both strings work in first overtone. That is i am not getting the root concept.
 
  • #12
199
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I am not understanding how will the both strings work in first overtone. That is i am not getting the root concept.
its not given that the second string is fixed at the other end. You just need to worry about the first string. If you are aware of boundary conditions, then you should not face any problem.
 
  • #13
301
7
What do we mean by light string??How will it interfere with the heavy string while making one node??
 
  • #14
199
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What do we mean by light string??How will it interfere with the heavy string while making one node??
light string means less linear mass density. The interference question is an interesting one, I'll suggest you read your textbook about this. It has to do with something with inertia.....let pulse travels through heavy string, when the pulse reaches the junction, two extreme conditions can be that the junction is a node or a antinode (intermediate conditions also happen). But when the pulse comes from the heavy string, it has more inertia compared to lighter string so it leads to an antinode in an extreme situation. The question you have given makes exactly this assumption to solve it.
 
  • #15
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  • #16
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So that means that the light string has no effect on the heavy string,isn't it?
 
  • #17
301
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Thanks for your precious guidance
 
  • #18
199
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So that means that the light string has no effect on the heavy string,isn't it?
yes, you can say that.
Thanks for your precious guidance
:thumbs:
 

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