Question

The displacement y is expressed as a function of x and t in terms of the wave numberk and the angular frequency. Write the equivalent equations in which y is shown as a function of x in terms of (a) kand v, (b) 1 and v(c) and fand (d) f and v.

Answer 1

Explanation:

The displacement equation is given by :

(a) $y=A\ sin(\omega t\pm kx)$

Since, $\omega=2\pi f$

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

$y=A\ sin(2\pi f t\pm kx)$

$y=A\ sin(\dfrac{2\pi vt}{2\pi /k}\pm kx)$

$y=A\ sin(kvt+kx)$

$y=A\ sink(vt+x)$

The above equation is in terms of k and v.

(b) $y=A\ sin(\omega t\pm kx)$

$y=A\ sin(2\pi \dfrac{v}{\lambda}t\pm\dfrac{2\pi}{\lambda}x)$

$y=Asin\dfrac{2\pi}{\lambda}(vt\pm x)$

(c) $y=A\ sin(\omega t\pm kx)$

$y=A\ sin(2\pi f t\pm \dfrac{2\pi}{\lambda} x)$

$y=A\ sin\ 2\pi (ft\pm \dfrac{x}{\lambda})$

(d) $y=A\ sin(\omega t\pm kx)$

$y=A\ sin(2\pi ft+\dfrac{2\pi}{\lambda}x)$

$y=A\ sin(2\pi ft+\dfrac{2\pi}{v/f}x)$

$y=A\ sin\ 2\pi f(t+\dfrac{x}{v})$

Where

k is the propagation constant

v is the speed of wave

f is the frequency of a wave

$\lambda$ is the wavelength

Hence, this is the required solution.