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The transverse displacement $y(x, t)$ of a wave on a string is given by $y(x, t)=e^{-\left(a x^2+b t^2+2 \sqrt{a b} x t\right)}$. This represents a
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The correct answer is:
wave moving in negative $x$-direction with speed $\sqrt{\frac{b}{a}}$
Phase of given wave is
$\phi=a x^2+b t^2+2 \sqrt{a b} x t$
Differentiating with respect to time,
$\frac{d \phi}{d t}=0=2 a x \frac{d x}{d t}+2 b t+2 \sqrt{a b}\left(t \cdot \frac{d x}{d t}+x\right)$
$\Rightarrow(2 a x+2 \sqrt{a b} t) \frac{d x}{d t}=-(2 \sqrt{a b} x+2 b t)$
So, wave velocity is
$v=\frac{d x}{d t}=\frac{-(2 \sqrt{a b} \cdot x+2 b t)}{(2 a x+2 \sqrt{a b} \cdot t)}=-\sqrt{\frac{b}{a}} \mathrm{~ms}^{-1}$
$\phi=a x^2+b t^2+2 \sqrt{a b} x t$
Differentiating with respect to time,
$\frac{d \phi}{d t}=0=2 a x \frac{d x}{d t}+2 b t+2 \sqrt{a b}\left(t \cdot \frac{d x}{d t}+x\right)$
$\Rightarrow(2 a x+2 \sqrt{a b} t) \frac{d x}{d t}=-(2 \sqrt{a b} x+2 b t)$
So, wave velocity is
$v=\frac{d x}{d t}=\frac{-(2 \sqrt{a b} \cdot x+2 b t)}{(2 a x+2 \sqrt{a b} \cdot t)}=-\sqrt{\frac{b}{a}} \mathrm{~ms}^{-1}$
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