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?
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 dont know how to calculate a value for ω with the given information