Question

A transverse wave is propagating along $+x$ direction. At $t=2s$, the particle at $x=4$ $m$ is at $y=2mm$. With the passage of time its $y$ coordinate increases and reaches to a maximum of $4mm$. The wave equation is (using $\omega$ and $k$ with their usual meanings)

• A. $y=4\sin \left(\omega (t+2)+k(x-2)+\frac{\pi }{6}\right]$
• B. $y=4\sin \left(\omega (t+2)+kx+\frac{\pi }{6}\right]$
• C. $y=4\sin \left(\omega (t-2)+k(x-4)+\frac{5\pi }{6}\right]$
• D. $y=4\sin \left(\omega (t-2)-k(x-4)+\frac{\pi }{6}\right]$

Answer 1

The correct option is D $y=4\sin \left(\omega (t-2)-k(x-4)+\frac{\pi }{6}\right]$

We know the general equation as
$y=Asin(\omega t-kx+\varphi )$
So as given at $t=2s$
$y=2mm$
$x=4mm$
Amplitude is given as $4mm$
SHM equation of the particle is
$\Rightarrow 2=4sin(2\omega -4k+\varphi )$
Solving this we get $\varphi =-2\omega +4k+\frac{\pi }{6}$
Putting this value in equation in wave equation
$\Rightarrow y=4sin(\omega t-kx+(-2\omega +4k+\frac{\pi }{6}$
$\Rightarrow y=4sin\left(\omega \right(t-2)-k(x-4)+\frac{\pi }{6}$