Standing wave transverse motion and amplitude

[Declan Purdy]

Homework Statement

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 0.381m. The maximum transverse acceleration of a point at the middle of the segment is 8600 m/^s and the max. transverse velocity is 3.4m/s.

What is the amplitude of the standing wave?

Homework Equations

y(x,t)=Asin(kx)sin(ωt)

The Attempt at a Solution

I calculated the wavelength of the fundamental frequency as 2L = 0.762

I then calculated k by 2π/λ = 8.246

I found the first partial derivative as
∂y(x,t)/∂t = ωAsin(kx)cos(ωt) = 3.4ms^-1

Then i found the second partial derivative of y(x,t) as
∂²y(x,t)/∂t² = -ω²Asin(kx)sin(ωt)Which is the same as -ω²y(x,t),
I know that y(x,t) will be the amplitide at the max acceleration, so 8600ms^-2 = -ω²A
I'm not sure where to go from here as I don't know how to calculate a value for ω with the given information

 

[mjc123]

What is the expression for the maximum velocity?

 

[Declan Purdy]

∂y/∂t = ωAsin(kx)cos(ωt) = 3.4ms^-1

 

[Chandra Prayaga]

The relevant equation you wrote is a traveling wave. The guitar string fixed at both ends exhibits standing waves.

 

[Declan Purdy]

The relevant equation you wrote is a traveling wave. The guitar string fixed at both ends exhibits standing waves.

Oh yes, thank you.
It is edited now but I am still at the same problem of solving for omega.

 

[Cutter Ketch]

You are only interested in the middle of the string where sin(kx) = 1, so you can get rid of that.

They give you the max velocity and the max acceleration. The maximums occur when the trig functions = 1, so you can get rid of those. Substitute in the given values and you will have two simple equations in two unknowns. Solve algebraically.