Question

An expression for the amplitude of a resultant standing wave on a vibrating string is 2Asinkx. Show that the maximum values of the amplitude of such wave at location x=(2n-1) times labda/2.

Answer 1

The extrema of function $f(x) = 2A\sin (kx)$ are located at the points:

$$f'(x_{e}) = 0\\ 2kA\cos (kx_e) = 0$$

The solutions of the last equation is:

$$x_e = \dfrac{\pi}{2k} + \dfrac{m\pi}{k},\space\space\space m = 0, \pm1, \pm2, ...$$

Since the wave is standing one, the locations of maxima and minima are the same. Thus, we can write for the maxima points:

$$x_m = \dfrac{\pi}{2k} + \dfrac{n\pi}{k},\space\space\space n = 0, \pm1, \pm2, ...\\ x_m =\dfrac{\pi(2n+1)}{2k}$$

Since

$$k = \dfrac{2\pi}{\lambda}$$

obtain:

$$x_m = \dfrac{(2n+1)\lambda }{4} = \left(2n+1\right)\dfrac{\lambda}{4}$$

Answer. The actual expression for the maxima is $x_m = \left(2n+1\right)\dfrac{\lambda}{4}$.