- #1

jimbo71

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## 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?