# 2 qn regarding wave

A 2 dimentional water wave spread in circular wave front.Show the amplitude A at a distance r from the initial diturbance is sq. rt. r.

i know that U is equal to 1/2 (mass/unit length)w^2A^2lumtha. How do i link the distance r to the eqn?

A horizontal string can transit a max of power, P with amplitude A and ang. frequency w.If the string is folded and used as a double string, tensoin is constant, what is the max. power?

So tension is doubled hence final Velocity is the square root of the initial velocity. However, i am not sure if angular freguency and amplitude will change.

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For question 1, I assume you mean inverse root r, since you need to show the result all that have to do is write down a solution, any solution that is circular, and show that the amplitude is what it is.

You may even be any to do it from dimensional analysis alone, you dont say want conditions for your waves are.

For question two I cant see what it is about is there more context?

what do you mean by solution?

for qn 2, the original qns is:
A horizontal string can transmit a max power of P if a wave with angularfrequency w and amplitude A is travelling along it. if the string is folded and use as a double string to increase its max power. What is the max. power that can be transmitted along the 'double string' if tension is constant?

HallsofIvy
Homework Helper
Wen said:
A 2 dimentional water wave spread in circular wave front.Show the amplitude A at a distance r from the initial diturbance is sq. rt. r.

i know that U is equal to 1/2 (mass/unit length)w^2A^2lumtha. How do i link the distance r to the eqn?
What is "U"? What is "lumtha"? (I suspect that was supposed to be "lambda" but you haven't told us what "lambda" means in this problem either.)

A horizontal string can transit a max of power, P with amplitude A and ang. frequency w.If the string is folded and used as a double string, tensoin is constant, what is the max. power?

So tension is doubled hence final Velocity is the square root of the initial velocity. However, i am not sure if angular freguency and amplitude will change.
Why do you say "So tension is doubled"? Doesn't the problem say "tension is constant"?? Looks to me like the linear density is doubled. What formula are you using?

I don't know what waves you have but in any radiating system the energy being radiated is being conserved so if you think about the expanding circle of "ripples" you will see immediately that the energy is going down as inverse the radius. If the system is linear then the energy is proprotional to the amplitude squared.

The above is an argument and you need a "show", so you only need to write down a solution that is circluar and "show" that it satisfies the wave equation you have.

"A solution" refers to a particular equation for the system in time and space. So for example you might have "The standing wave solution" or the "Planar wave solution" or the "Cylindrical wave solution" for the system.

For example if I was asked to show that the amplitude of a plane wave in particular system was constant, I would write down F=A cos (wt+kx), then write down the wave equation for the system in terms of differentials in time and space for F. Then substituting my proposed solution into LHS and RHS of the wave equaiton I would find that A cancled out, thus showing it to be a constant. "F" could be the electric field in maxwels equaiton or it could be air pressure in sound waves, the solution could be sin(wt+kx) or e^i(wt+kx). it could be a linear combination of w1,k1,w2,k2 and so on, but obviously that would be silly.

You should have access to or be able to find the cylindrical/circular wave solution that applies to the system you are being asked about. You only need to "find" the solution in the sense of presenting it from having looked it up in a book (not in the sense of having derived it from the equations). This solution is effectively (by definition) a solution to the wave and has a term inverse root radius in it.

okay i see. ya, its the linear intensity that doubles. I think that amplitude and angular frequency will not change as its dependant upon the source of vibration right?

Thanks gnpatterson for your explanation. I just started learning on waves and have not reached the sub-chapter on spherical and planes, hence i wasn't aware that the answer is so direct.