Travelling Wave Equation

A travelling wave written as $y=A\sin (kx-\omega t)$ has wave number $k$ and angular frequency $\omega$, and propagates at speed $v=\omega /k$. This ratio measures how fast a point of constant phase advances along the medium. The minus sign separating the space and time terms also tells us the wave travels in the positive x direction. Comparing with the given form, $k=4\pi \text{ rad/m}$ and $\omega =200\pi \text{ rad/s}$. Therefore $v=\frac{200\pi }{4\pi }=50\text{ m/s}$. The value $200\text{ m/s}$ wrongly reports $\omega$ alone as if it were a speed, ignoring the wave number. The value $25\text{ m/s}$ halves the correct ratio, a slip from misreading $k$ as $8\pi$. The value $100\text{ m/s}$ comes from using only half of $\omega$ over $k$. This applies the standard NCERT relation that wave speed equals angular frequency divided by wave number, equivalently $v=f\lambda$. As a plausibility check, the wavelength here is $\lambda =2\pi /k=0.5\text{ m}$ and the frequency is $f=\omega /2\pi =100\text{ Hz}$, giving $v=f\lambda =100\times 0.5=50\text{ m/s}$, which confirms the result independently.