Normal mode of string in third harmonic

[jimbo71]

Homework Statement

A string that is fixed at both ends is vibrating in the third harmonic. the wave has a speed of 186m/s and a frequency of 225Hz. the amplitude of the string at an antinode is 0.0037m.

How much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point?

Homework Equations

y(x,t)=(Aswsin(kx)(sin(omega*t))

The Attempt at a Solution

I think there is a node at x=0 and the third harmonic means there are three antinodes between x=0 and x=L. Therefore the antinodes are located at x=1/6L,1/2L,5/6L and the nodes are located at x=0,2/6L,4/6L,L. I set 0.0037m=y(1/6L,t)=(Aswsin(kx))(sin(omega*t)) and solve for t where
L=186/225*3/2
Asw=0.0037
k=(2pi*225)/186
x=1/6L
omega=2pi*225
I found the first antinode to occur at t=1/900s
However when I solve for t with x=2/6L I cannot calculate t because it requires me to divide by zero because sin(k*2/6L)=0
what am i doing wrong?

 

[Prosthetic Head]

This is just a thought that might simplify things a lot, the frequency of the wave describes how many times the wave passes a point per second, in this case an antinode. Also, each time the wave passes an antinode it has completed one full cycle from peak to trough and back to peak again.

 

[thomate1]

Your approach to finding the time it takes for the string to go from its largest upward displacement to its largest downward displacement at the antinode is correct. However, the issue with dividing by zero is due to a mistake in your calculation of the second antinode location. The second antinode should be located at x=3/6L, not 2/6L. This means that your calculation for t with x=2/6L should be t=1/450s, not t=1/900s. This correction should allow you to continue with your solution and find the correct time for the string to go from its largest upward displacement to its largest downward displacement at the antinode. Keep up the good work in solving these types of problems!