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arindameducationusc

  • one year ago

had a small question, how to take out transverse speed of a wave @Irishboy123

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  1. arindameducationusc
    • one year ago
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    See attachment, only part b needed

  2. arindameducationusc
    • one year ago
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    Check point 2

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  3. arindameducationusc
    • one year ago
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    @Michele_Laino

  4. IrishBoy123
    • one year ago
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    the transverse speed is the speed perpendicular to wave travel and technically is the partial derivative wrt t so for \(y(x,y) = 2 \sin (4x - 2t)\) \(v _{\perp} = \dfrac{\partial }{\partial t}(2 \sin (4x - 2t))\) you can get the same result just by choosing a fixed point on the x axis, say , x = 0 and looking at \(\dfrac{d }{d t}(2 \sin (- 2t)\) instead. as if it were a standing wave. at all points you will get the same oscillation in the transverse direction so this works nicely so that leaves us with \(v_{\perp}= -4 \cos(-2t) = -4 \cos(2t)\), and we have a wave whose velocity transverse alternates between 4 and -4 as cosine alternates between -1 and 1. so max speed for that = 4 repeat for others....

  5. arindameducationusc
    • one year ago
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    Got it @IrishBoy123 Thank you

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