Difficult but interesting standing waves question

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Homework Help Overview

The discussion revolves around a physics problem involving standing waves on a rope system consisting of two strings with different linear mass densities. The original poster seeks to determine the wavelength of the first overtone while grappling with the implications of having two strings connected at one end.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the boundary conditions relevant to the problem, such as the uniform tension and frequency across both strings. There are inquiries about how the differing mass densities of the strings affect the formation of nodes and antinodes in the standing wave pattern.

Discussion Status

The conversation includes attempts to clarify the role of the light string in relation to the heavy string, with some participants suggesting that the junction behaves as an antinode. There is acknowledgment of the complexity of the problem, and while some participants express confusion, others provide insights into the underlying physics concepts.

Contextual Notes

Participants note the lack of explicit information regarding the boundary conditions of the second string and the implications of this on the overall understanding of the problem. The discussion reflects a mix of personal experiences and academic preparation, particularly in relation to competitive exams.

nil1996
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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
 
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nil1996 said:
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!
 
NihalSh said:
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.
 
nil1996 said:
Nihal are you from india.I am from india.

Well, yes I am. But that is irrelevant to the question.:wink:
 
NihalSh said:
Well, yes I am. But that is irrelevant to the question.:wink:

are you preparing for iitjee??
 
nil1996 said:
are you preparing for iitjee??

Why? Are you?
 
NihalSh said:
Why? Are you?

Yes i am preparing for iitjee without coaching.
 
nil1996 said:
Yes i am preparing for iitjee without coaching.

good luck. If you want talk about iitjee and all that, use private message.
 
Still not getting the problem
 
  • #10
nil1996 said:
Still not getting the problem

What is it that you are not getting?
 
  • #11
I am not understanding how will the both strings work in first overtone. That is i am not getting the root concept.
 
  • #12
nil1996 said:
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
What do we mean by light string??How will it interfere with the heavy string while making one node??
 
  • #14
nil1996 said:
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.
 
  • #16
So that means that the light string has no effect on the heavy string,isn't it?
 
  • #17
Thanks for your precious guidance
 
  • #18
nil1996 said:
So that means that the light string has no effect on the heavy string,isn't it?
yes, you can say that.
nil1996 said:
Thanks for your precious guidance

:thumbs:
 

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