# Difficult but interesting standing waves question

nil1996

## 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

NihalSh
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!!!

nil1996
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.

NihalSh
Nihal are you from india.I am from india.

Well, yes I am. But that is irrelevant to the question.

nil1996
Well, yes I am. But that is irrelevant to the question.

are you preparing for iitjee??

NihalSh
are you preparing for iitjee??

Why? Are you?

nil1996
Why? Are you?

Yes i am preparing for iitjee without coaching.

NihalSh
Yes i am preparing for iitjee without coaching.

good luck. If you want talk about iitjee and all that, use private message.

nil1996
Still not getting the problem

NihalSh
Still not getting the problem

What is it that you are not getting???

nil1996
I am not understanding how will the both strings work in first overtone. That is i am not getting the root concept.

NihalSh
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.

nil1996
What do we mean by light string??How will it interfere with the heavy string while making one node??

NihalSh
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.

nil1996
So that means that the light string has no effect on the heavy string,isn't it?

nil1996