Stationary waves in composite strings

[arvindsharma]

Dear All,

I was reading the concept of stationary waves in composite string ABC made up of joining two strings AB AND BC with different mass per unit length and a joint at 'B'.the two ends of the composite string are clamped at 'A' and 'C' and a transverse wave is set up by an external source at one of the clamp say clamp 'A'.it was written in the book that stationary wave will set up in the composite string ABC at a particular frequency of external source with joint 'B' as a node.following are my doubts
1)the stationary wave in AB must be due to incident wave and reflected wave at junction B but at junction 'B' there is some reflection and some transmission(so that stationary wave in B can also be set up).due to partial reflection the amplitude of incident wave and reflected wave must be different so a perfect stationary wave is not possible in string AB.however wave in string BC is perfectly stationary(assuming no energy loss at clamp C).is my reasoning correct in this sense?
2)Why there is a node at junction?does this always happen in composite string that joint is a node?
3)the stationary wave in string AB and BC will have different amplitude because the amplitude of reflected and transmitted wave is not same?

 

[Simon Bridge]

1. because of the partial refection at B you won't get opposite traveling waves to interfere to produce a non-zero amplitude there and get standing waves. You can check by solving the wave equation for standing waves or just do the experiment.
2. see above. yes - for standing waves.
3. yes

Merry Xmas.

 

[AlephZero]

THinking about a mode of vibration as a "stationary wave" made from "traveling waves" in opposite directions is useful for a simple structure like a uniform string, but less useful for more complicated structures.

You will certainly get "stationary" modes of vibration of the whole structure, but the traveling waves are complicated because the speed will be different in the two parts of the string, and there will be partial reflection and partial transmission at B, for the traveling waves going in both directions. The "shape" of the traveling waves will not be the same as the shape of the stationary vibration mode, and the "shape" of the stationary wave willl not be a simple sine wave like a uniform string

In general, point B will not be a node. If it is a node in your example, there must be something special about the problem, for example some relation between the mass per unit length of the two strings, or the lengths of parts AB and BC. Or, the book is wrong!

 

[arvindsharma]

not yet satisfied.please clear my doubts

 

[AlephZero]

You need to tell us EXACTLY what the book says about the system.

it should be obvious that if AB and BC were the same length, and the mass of BC was 1.01 times the mass of AB, the vibration modes of the system would be very close to a uniform string, and there would be standing waves that did NOT have a node at B.

If the system in the book has a node at B, it is because of the particular masses, lengths, etc, of that system, not because every possible set of two strings has a node there.

 

[sophiecentaur]

You need to tell us EXACTLY what the book says about the system.

it should be obvious that if AB and BC were the same length, and the mass of BC was 1.01 times the mass of AB, the vibration modes of the system would be very close to a uniform string, and there would be standing waves that did NOT have a node at B.

If the system in the book has a node at B, it is because of the particular masses, lengths, etc, of that system, not because every possible set of two strings has a node there.

Of course. To get a standing wave, you must have solutions to the wave equations on both sections that are satisfied by the boundary conditions. I have a feeling that the bandwidth of the system (i.e. the losses) may allow some latitude but I can't be sure.

 

[AlephZero]

To get a standing wave, you must have solutions to the wave equations on both sections that are satisfied by the boundary conditions. I have a feeling that the bandwidth of the system (i.e. the losses) may allow some latitude but I can't be sure.

You are right, the existence of "exact" standing waves with "perfect" nodes (zero motion at all times) is only an approximation, and depends on the way the system is damped.

Making models of damping based on physics is notoriously difficult, especially for lightly damped systems like vibrating strings etc. The usual technique is to make a model that is mathematically convenient, and fits reasonably well to measurements close to the resonant frequency of each mode. What happens away the resonances often isn't important from an engineering point of view, because the amplitudes are small compared with the resonance peaks.

In real life, different positions on a vibrating object don't necessarily vibrate in phase with each other, and don't necessarily even vibrate back and forth through some equilibrium position. For example it a rotating object is vibrating, the coriolis forces make every point move in an ellipse around the equilibrium point (as viewed by somebody rotating at the same speed as the object), not back and forth along a straight line. And the orientations of the principal axes of the ellipses vary at different positions on the object.

But all this is probably irrelevant to the OP's question.

 

[arvindsharma]

i got the point. thank you all for your contribution