The transverse displacement y(x, t) of a wave on a string is given by y(x, t)=e^ -(a x^ 2 +b t^ 2 +2 a b . xt ) This represents a

Correct answer: A. wave moving in -x direction, speed (b)/(a) y(x, t)=e^ -(a x^ 2 +b t^ 2 +2 a b x t) =e^ -( a x + b t )^ 2 It is a function of type y=f(ω t+k x) y(x, t) represents wave travelling along -x direction. Speed of wave =(ω)/(k)= b a = (b)/(a)

BITSAT 2012

Exam: BITSAT
Year: 2012
Subject: Physics
Chapter: Waves and Sound

BITSAT 2012 · Physics · Waves and Sound

The transverse displacement y(x, t) of a wave on a string is given...

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}\right.}

)

This represents a

Options

Awave moving in $-x$ direction, speed $\sqrt{\frac{b}{a}}$
Bstanding wave of frequency $\sqrt{b}$
Cstanding wave of frequency $\frac{1}{\sqrt{b}}$
Dwave moving in $+x$ direction, speed $\sqrt{\frac{a}{b}}$

Correct answer

A. wave moving in $-x$ direction, speed $\sqrt{\frac{b}{a}}$

Step-by-step solution

y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}=e^{-(\sqrt{a x}+\sqrt{b t})^{2}}

It is a function of type

y=f(\omega t+k x)\therefore y(x, t)

represents wave travelling along

-x

direction.

\text{ Speed of wave }=\frac{\omega}{k}=\frac{\sqrt{b}}{\sqrt{a}}=\sqrt{\frac{b}{a}}

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