Question

A string of length is vibrating in its 3rd overtone, with maximum amplitude A. The corresponding wave equation is

Answer 1

Given:

A string of length is vibrating in its 3rd overtone, with maximum amplitude A.

To find:

Wave equation

Calculation:

Since the string is vibrating in its 3rd overtone we can say that a standing wave has been formed.

Refer to the attached diagram.

In the 3rd overtone , nodes will be formed at

x = 0 , L/4 , 2L/4 , 3L/4 and L

Generalized equation of standing wave is :

$y = (max \: amp) \bigg \{ \sin( \omega t) \cos(kx) \bigg \}$

Now , from the diagram , we know that :

$\therefore \: 4 \times \dfrac{ \lambda}{2} = L$

$= > \: \lambda = \dfrac{ L }{2}$

Continuing with wave equation :

$y = (max \: amp) \bigg \{ \sin( \omega t) \cos \{ (\frac{2\pi}{ \lambda}) x \} \bigg \}$

$= > y =a\bigg \{ \sin( \omega t) \cos \{ (\frac{2\pi}{ \lambda}) x \} \bigg \}$

$= > y =a\bigg \{ \sin( \omega t) \cos \{ (\frac{2\pi}{ ( \frac{L}{2}) }) x \} \bigg \}$

$= > y =a\bigg \{ \sin( \omega t) \cos \{ (\frac{4\pi}{L }) x \} \bigg \}$

$= > y =a\bigg \{ \sin( \omega t) \cos ( \frac{4\pi x}{L } ) \bigg \}$

So , final answer is :

$\boxed{ \bold{ \red{y =a\bigg \{ \sin( \omega t) \cos ( \frac{4\pi x}{L } ) \bigg \}}}}$