Travelling Wave Equation

Question

A transverse wave on a stretched rope is described by y equals 0.02 sin of the quantity 4 pi x minus 200 pi t, with all quantities in SI units. What is the speed at which this wave travels along the rope?

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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, $k = 4\pi	ext{ rad/m}$ and $\omega = 200\pi	ext{ rad/s}$. Therefore $v = \frac{200\pi}{4\pi} = 50	ext{ m/s}$. The value $200	ext{ m/s}$ wrongly reports $\omega$ alone as if it were a speed, ignoring the wave number. The value $25	ext{ m/s}$ halves the correct ratio, a slip from misreading $k$ as $8\pi$. The value $100	ext{ m/s}$ 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 $\lambda = 2\pi/k = 0.5	ext{ m}$ and the frequency is $f = \omega/2\pi = 100	ext{ Hz}$, giving $v = f\lambda = 100 	imes 0.5 = 50	ext{ m/s}$, which confirms the result independently.

Travelling Wave Equation

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