The transverse displacement y(x, t) of a wave | vedclass

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(D) The given equation is $y(x, t) = e^{-(ax^2 + bt^2 + 2\sqrt{ab}xt)}$.We can rewrite the exponent as a perfect square: $ax^2 + bt^2 + 2\sqrt{ab}xt =

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a wave moving in the $-x$ direction with speed $\sqrt{\frac{b}{a}}$

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(D) The given equation is $y(x, t) = e^{-(ax^2 + bt^2 + 2\sqrt{ab}xt)}$.We can rewrite the exponent as a perfect square: $ax^2 + bt^2 + 2\sqrt{ab}xt = (\sqrt{a}x + \sqrt{b}t)^2$.Therefore,the wave function is $y(x, t) = e^{-(\sqrt{a}x + \sqrt{b}t)^2}$.$A$ general traveling wave function is represented as $f(kx + \omega t)$,where the wave moves in the $-x$ direction if the sign between $kx$ and $\omega t$ is positive.Comparing $y(x, t) = e^{-(\sqrt{a}x + \sqrt{b}t)^2}$ with $f(kx + \omega t)$,we identify $k = \sqrt{a}$ and $\omega = \sqrt{b}$.The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{\sqrt{b}}{\sqrt{a}} = \sqrt{\frac{b}{a}}$.Since the expression is of the form $f(kx + \omega t)$,the wave is moving in the $-x$ direction.

The transverse displacement $y(x, t)$ of a wave on a string is given by $y(x, t) = e^{-(ax^2 + bt^2 + 2\sqrt{ab}xt)}$. This represents:

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