WBJEE 2023 Physics & Chemistry Solved Paper

1. The general equation of a wave propagating in one dimension can be expressed as $y=A \sin (\omega t - k x + \varphi)$, where $A$ is the amplitude, $\omega$ is the angular frequency, $k$ is the wave number, $x$ is the position, $t$ is the time, and $\varphi$ is the phase constant.2. The given wave equation is $y=3 \sin 2\pi \left(-\frac{x}{3}\right)$. This can be rewritten as $y=3 \sin \left(-\frac{2\pi x}{3}\right)$, which matches the form $y=A \sin (\omega t - k x)$ where $\omega t$ is zero (i.e., $t=0), A$ is 3 and $k$ is $-\frac{2\pi}{3}$.3. The negative sign of $k$ indicates that the wave is moving in the negative $x$-direction, but the problem states that the wave is moving in the positive $x$-direction. Hence, we should take $k$ as positive, i.e., $k=\frac{2\pi}{3}$.4. The speed of the wave $v$ is given by $v=\frac{\omega}{k}$. From the problem, we know that $v=4 \text{ m/s}$. By substituting these values, we can solve for $\omega$ : $4=\frac{\omega}{\frac{2\pi}{3}}$ which gives $\omega=\frac{8\pi}{3}$.5. So, the wave equation at any time $t$ can be written as $y=3 \sin \left[\frac{8\pi}{3} t - \frac{2\pi}{3} x\right]$.6. Now, we want to find the wave equation at $t=4 \text{ s}$. Substituting $t=4$ into the equation gives $y=3 \sin \left[\frac{8\pi}{3} \times 4 - \frac{2\pi}{3} x\right]$.7. Simplifying the argument of the sine function, we get $y=3 \sin 2\pi \left(\frac{-x+16}{3}\right)$.8. Rearranging the argument of the sine function again, we get $y=3 \sin 2\pi \left(-\frac{x-16}{3}\right)$.