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

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(B) The given wave 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}

Vedclass · Accepted Answer

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

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(B) The given wave equation is $y(x, t) = e^{-(ax^2 + bt^2 + 2\sqrt{ab}xt)}$.We can rewrite the exponent as a perfect square:$ax^2 + bt^2 + 2\sqrt{ab}xt = (\sqrt{a}x + \sqrt{b}t)^2 = a(x + \sqrt{\frac{b}{a}}t)^2$.Thus,$y(x, t) = e^{-a(x + \sqrt{\frac{b}{a}}t)^2}$.This is a function of the form $y = f(x + vt)$,which represents a wave traveling in the negative $x$-direction.Comparing $x + \sqrt{\frac{b}{a}}t$ with $x + vt$,we get the wave speed $v = \sqrt{\frac{b}{a}}$.Therefore,it represents a wave moving in the $-x$ direction with speed $\sqrt{\frac{b}{a}}$.

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 a:

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