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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $\sqrt{\frac{b}{a}}$

Vedclass · Answer

(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:

Similar Questions

The wave equation is $y = 0.30 \sin (314t - 1.57x)$ where $t, x$ and $y$ are in second,meter and centimeter respectively. The speed of the wave is ..... $m/s$.

The equation of a wave is represented by $y = 10^{-4} \sin(100t - x/10)$. The velocity of the wave will be .... $m/s$.

$A$ wave is given by $y = 3 \sin 2\pi \left( \frac{t}{0.04} - \frac{x}{0.01} \right)$,where $y$ is in $cm$. The frequency of the wave and the maximum acceleration of the particle are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module