## wave on a guitar string

A guitar string is vibrating in its fundamental mode, with nodes at each end. The length of the segment of the string that is free to vibrate is L. The maximum transverse acceleration of a point at the middle of the segment is a and the maximum transverse velocity is v.

What is the amplitude of this standing wave?

I worked out the fundamental frequency, which I think is v/2L, but I'm not sure how to get A, since all of the formulae that I know relating to A involve k.

Any help would be great.

Thanks.

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 Have a look at the equations for the transverse velocity and acceleration of the wave: $\frac{\partial y}{\partial t} = \omega A\sin(kx - \omega t)$ $\frac{\partial^2y}{\partial t^2} = -\omega^2A\cos(kx - \omega t)$. What are the maximum values these quantities can assume?
 Ok, as the guitar string is fixed on both sides . its wavelength(Y) is related to length as: $L= \frac{nY}{2}$ For fundamental frequency , n=1 Therefore, $L= Y/2$ Now max. transverse acceleration is given by: $a=Aw^2$ you know a and w (because you know frequency) Calculate A.

## wave on a guitar string

I ended up with aL^2/(pi^2*v^2), but it states that it does not depend on the variable L.

 Thats ok... i found it
 Care to elaborate as to how to solve this problem?