Question

A wave is represented by the equation $y = \left[ {A\sin \left( {10x + 15t + \dfrac{1}{3}} \right)} \right]$ where $x$ is in meters and $t$ is in seconds. The expression represents …...
(A) A wave traveling in positive $x - direction$ with a velocity $1.5m{s^{ - 1}}$
(B) A wave traveling in negative $x - direction$ with a velocity $1.5m{s^{ - 1}}$
(C) A wave traveling in negative $x - direction$ with a wavelength $2\;m$
(D) A wave traveling in positive $x - direction$ with a wavelength $2\;m$

Answer 1

Hint : We can initiate the sum by comparing the given equation with the standard equation of wave and noting the given values. The required values can be obtained from the given values. The direction of the wave depends on the sign of the angular velocity term.

Complete Step By Step Answer:
Let us first consider a sinusoidal wave function
$y = A\sin \left( {kx + \omega t + \phi } \right)$
By comparing the standard equation with the given equation we can get the values as
$k$ = Wave Number = $\dfrac{{2\pi }}{\lambda } = 10$
$\omega$ = Angular velocity = $\dfrac{{2\pi }}{T} = 15$
$\phi$ = Phase = $\dfrac{1}{3}$
$\lambda$ = Wavelength
$T$ = Time period
Now, for wavelength from the formula of wavenumber,
$\therefore \lambda = \dfrac{{2\pi }}{k} = \dfrac{{2\pi }}{{10}}$
$\therefore \lambda = 0.628m$
Now, we know that velocity is displacement per unit time.
For a wave, velocity can be defined as the wavelength of the wave per time period of one wave.
$\therefore v = \dfrac{\lambda }{T}$
But we know $\lambda = \dfrac{{2\pi }}{k}$
$\therefore v = \dfrac{{2\pi }}{{k \times T}}$
Also from the formula of angular velocity $\omega = \dfrac{{2\pi }}{T}$
$\therefore v = \dfrac{\omega }{k}$
Substituting the values from the given data,
$\therefore v = \dfrac{{15}}{{10}}$
$\therefore v = 1.5m{s^{ - 1}}$
For the direction of the propagation of the wave, let us consider a crest where $\sin \theta = 1$ and the initial phase to be $0$ .
Thus we get $\sin \left( {10x + 15t} \right) = 1$
$10x + 15t = \dfrac{\pi }{2}$
Now, as the wave propagates, the time $t$ increases, and with time, if the distance $x$ also increases then the phase does not remain constant.
Hence, to maintain the phase at $\dfrac{\pi }{2}$ , the distance $x$ must be negative, which proves that the wave is traveling in the negative $x - direction$ .
Therefore the wave is traveling in the negative $x - direction$ with velocity $1.5m{s^{ - 1}}$
The correct solution is Option $(B)$ .

Note :
The direction of the wave depends on the sign of the wavenumber term, but indirectly it depends on the sign of the angular velocity term. If the sign is positive, then the wave moves in negative $x - direction$ , whereas if the sign before angular velocity term is negative, the wave moves in the positive $x - direction$ .