Amplitude of Standing Wave in Fundamental Mode

[Thomas_]

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 8200 m/s^2 and the max. transverse velocity is 4m/s.

What is the amplitude of the standing wave?

Homework Equations

y(x,t)=Asin(kx)sin(\omega t)

The Attempt at a Solution

I figured that the wavelength in fundamental mode is given by \lambda = 2L = 0.762. Then k is k=\frac{2\pi}{\lambda}.
The partial der. of the wave equation are:

\frac{\partial y}{\partial t}=\omega Asin(kx)cos(\omega t) = 4
\frac{\partial^2 y}{\partial t^2}=- \omega^2 Asin(kx)sin(\omega t) = 8200I'm not sure what to do next. The time variable in the equation is confusing me, there are 3 unknowns (omega, t, A) but I only have these two equations.

 

[berkeman]

Not sure you are differentiating correctly, at least for the 2nd derivative. Also, what can you say about the times t with respect to the overall period T where the max velocity and max acceleration occur. There not at the same t, right?

 

[michalll]

When its said maximum, it means the sines and cosines are =1.

 

[Thomas_]

Ups sorry, the differentiation error was just a typing mistake ;)

Also, what can you say about the times t with respect to the overall period T where the max velocity and max acceleration occur. There not at the same t, right?

When the velocity is maximum, the acceleration at that point should be 0? So it can't be the same t. But how does that help me?

 

[berkeman]

When the velocity is maximum, the acceleration at that point should be 0? So it can't be the same t. But how does that help me?

See michalll's hint...