Chapter 11, Problem 81GP

To determine

The wavelength of the resonant vibration.

Answer to Problem 81GP

The wave length of the resonant vibration $\frac{4L}{2n-1}$ .

Explanation of Solution

Introduction:

A string has a free end = antinode

String has a fixed end = node

Here minimum distance from a node to antinodes is used.

Maximum distance from a node to antinodes is

  $\frac{\lambda }{4},\frac{3\lambda }{4}\frac{5\lambda }{4}$

So, the general relation is

  $\frac{(2n-1)\lambda }{4}$ , $n=1,2,3,4$ …….

So, $\lambda =\frac{4L}{2n-1}$

Conclusion:

The wavelength of the resonant vibration is $\frac{4L}{2n-1}$ .