Reading a progressive wave equation: how do I find direction, speed, wavelength, and what is particle speed?

6 answers & replies

• [+][–]Kabir Joshi11thSample· 9d ago

  Direction rule: look at the signs of the $x$ and $t$ terms inside the sine. Opposite signs ($kx - \omega t$ or $\omega t - kx$) mean the wave travels in $+x$. Same signs ($kx + \omega t$) mean it travels in $-x$. So yours is $+x$ direction; flipping the order only changes the overall sign of the whole function, like a phase shift of $\pi$, it doesn't change direction.

  Then: $v = \omega/k = 100/5 = 20$ m/s, $\lambda = 2\pi/k = 2\pi/5 \approx 1.26$ m, $f = \omega/2\pi \approx 15.9$ Hz. Quick check: $f\lambda = 15.9 \times 1.26 \approx 20$. ✓

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• [+][–]Yash Tiwari11thSample· 9d ago

  particle max speed = A ω = 0.02 × 100 = 2 m/s so same as wave speed na? both are v... Diya just use that bhai. Meri notebook me bhi likha hai 'v = ω/k = Aω' aise hi, so mujhe to same lagta hai

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  • [+][–]Arjun Nair12thSample· 9d ago

    No, they are two different speeds that happen to have the same symbol. The wave speed $v = \omega/k = 20$ m/s is how fast the pattern moves along $x$. The particle just goes up and down about its mean position, so its speed is $\partial y/\partial t = 0.02\times100\cos(100t-5x)$, which peaks at $A\omega = 2$ m/s. Here 20 and 2 are different. Don't equate them. (For a wave, they would be equal only by coincidence.)

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    • [+][–]Diya Menon11thSample· 9d agoAuthor

      OMG ok that makes sense, a pattern moving vs the actual particle wiggling. Thank you Arjun and Kabir, my tiny brain is saved 🫶

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• [+][–]Riya Singh12thSample· 9d ago

  Phase difference, I like to picture it as a hill moving. A point further along $x$ sees the same crest later. Over one full wavelength $\lambda$ the phase changes by $2\pi$, so

  $\Delta\phi = k\,\Delta x = \frac{2\pi}{\lambda}\Delta x$

  For your wave, $\lambda \approx 1.26$ m. Two points $0.314$ m apart, a quarter of a wavelength, differ in phase by $\pi/2$. For the same point at two different times, $\Delta\phi = \omega\,\Delta t$. Time and space give the same idea in two units.

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• [+][–]Siddharth RaoDropperSample· 9d ago

  Exam trick: always compare to the standard form $y = A\sin(\omega t - kx)$ and read off $A$, $\omega$, $k$. Everything else is a one-line formula from those three. If the equation has a messy form like $y = 0.02\sin(200\pi t - 10\pi x)$ you can simplify: $\omega = 200\pi$, $k = 10\pi$, $v = 20$ m/s, $f = 100$ Hz, $\lambda = 0.2$ m. Also if the equation is given with a number outside, like $\sin(2(t - x/10))$, expand first and then read off.

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