Question

A transverse wave is propagating along $+x$ direction. At $t=2s,$ the particle at $x=4m$ 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. $y=4\sin \left[\omega \right(t-2)-k(x-4)+\frac{\pi }{6}]$
• B. $y=4\sin \left[\omega \right(t+2)+k(x)+\frac{\pi }{6}]$
• C. $y=4\sin \left[\omega \right(t-2)-k(x-4)+\frac{5\pi }{6}]$
• D. $y=4\sin \left[\omega \right(t+2)+k(x-2)+\frac{\pi }{6}]$

Answer 1

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

As we know, general equation of wave is given by,
$y=4\sin (\omega t+kx+\varphi )$

Hence, according to the given question, 𝑆𝐻𝑀 equation of the particle at $x=4$
$y=4\sin \left[\omega \right(t+2)+\frac{\pi }{6}]$

Therefore,
Wave equation replacing $t$ by $[t-\left(\frac{x-4}{v}\right)]$ is
$y=4\sin [\omega (t-\frac{(x-4)}{v}-2)+\frac{\pi }{6}]$
$y=4\sin \left[\omega \right(t-2)-\frac{\omega }{v}(x-4\left)\frac{\pi }{6}\right]$
$y=4\sin \left[\omega \right(t-2)-k(x-4\left)\frac{\pi }{6}\right]$

Final answer: $\left(d\right)$