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We have a composite string of two materials A and B which have been joined end-to-end to form the string. The free ends are clamped to two supports and the system looks like this: The Setup

Objective: Find $\lambda$ of the standing wave generated on this string

Approach:

Now if a wave is generated here we have two nodes at the fixed (clamped) ends and again we have a node at the joint b/w A and B. So at first I tried to find out the $\lambda$ (wavelength) for the wave, but I am facing a major problem.

The distance b/w the three consecutive nodes are not the same. So can such a wave be generated? please help me.

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There is no reason for the junction between the two parts of the string to be a node.

Now if we can assume the tension of the string is constant, and the mass per unit length of the string is $\rho_A$ , $\rho_B$ respectively, then we know the ratio of wavelengths in each section

$$\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{\rho_B}{\rho_A}}\tag1$$

Now an integer number of half-wavelengths requires that

$$\frac{L_A}{\lambda_A}+\frac{L_B}{\lambda_B} = \frac12 n\tag2$$

where $n$ is an integer.

This is something you should now be able to solve by expressing $\lambda_B$ in terms of $\lambda_A$ (from (1)) and substituting into (2).

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  • $\begingroup$ Thanks for the answer. I understood the approach but the problem is that the only data given in the question are the lengths of the strings A and B, even nothing is said about their densities. $\endgroup$ – jyoti proy Apr 5 '17 at 8:48
  • $\begingroup$ One could say that the wavelength is just twice the distance between the two end points without worrying about the fact that the wavelength is different in the two parts of the string. Not very satisfying but perhaps that is the "answer "... $\endgroup$ – Floris Apr 5 '17 at 10:41
  • $\begingroup$ The book contains a solution which I could not understand. It takes the HCF (or GCD) of the two lengths given(60 cm and 40 cm) which is 20 and treats it as a common node-node distance of the two. Thus it says $\frac{\lambda}{2}$ =20 thus $\lambda$ =40. I just couldn't get it why they are taking the HCF. $\endgroup$ – jyoti proy Apr 5 '17 at 19:15
  • $\begingroup$ OK - what they are saying is "the junction is a node" (although there is no reason why it would be). Once you decide that, then you need a (half)-wavelength that fits in both A and B. This means that an integer number of half-wavelengths needs to fit in both 40 cm and 60 cm. The GCD of 20 is the biggest half-wavelength that would fit. $\endgroup$ – Floris Apr 5 '17 at 19:21
  • $\begingroup$ Unfortunately, it's not at all obvious to me why there would be a node (unless that is stipulated). And if there is a node, then that tells you something about the density of the two parts of the string. A very confusion question - but I'm glad I was able to clarify things at least a little bit for you. $\endgroup$ – Floris Apr 5 '17 at 21:00
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At a particular frequency you need to fit an integer number of half wavelengths between each of the parts of the string between the two nodes.
Since the wavelength will differ so will the number of half wavelength on either side of the join between the two types of string.

If you had just string $A$ then you know of a formula which links the harmonic frequencies $f_{n_{\rm A}}$ of that length of string $l_{\rm A}$ to the tension in the string $T$ , the mass per unit length of the string $\mu_{\rm A}$ and an integer which is the label for the harmonic $n_{\rm A}$.
The same is true of string $B$.
You are looking for the frequency of oscillation to be the same for each string which has the same tension with the condition that $n_\rm A$ and $n_\rm B$ are integers.

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  • $\begingroup$ Thanks for the answer. It helped a lot, but the problem is that only the length of string A and B is given, nothing is mentioned about their densities. $\endgroup$ – jyoti proy Apr 5 '17 at 8:50

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