Question

Equation of a travelling wave is $ y(x,t) = 5{e^{ - (a{x^2} + b{t^2} + 2xt\sqrt {ab} )}} $ where x & y is in metres and t is in seconds. $ a = 25{m^{ - 2}} $ And $ b = 9{\sec ^{ - 2}} $
[A] Travelling wave is propagated along $ ( + ) $ x direction
[B] Travelling wave is propagated along $ ( - ) $ x direction
[C] Speed of wave is $ \dfrac{3}{5}m{\sec ^{ - 1}} $
[D] Maximum displacement of particle is $ 5m $

Answer 1

Hint :In order to solve this question we will use the definition of wave which states that it is a disturbance in a medium which carries energy. It propagates both in space and time. In general it is of the form of sine or cosine which can be expressed in exponential format also. Its direction can be found using gradients and depends on the sign between x and t.

Complete Step By Step Answer:
Consider wave equation defined in question as $ y(x,t) = 5{e^{ - (a{x^2} + b{t^2} + 2xt\sqrt {ab} )}} $
We will first simplify this wave equation as $ y = 5{e^{ - \{ {{(x\sqrt a + t\sqrt b )}^2}\} }} $
 $ \Rightarrow y = 5{e^{ - 2}} \times {e^{(\sqrt a x + \sqrt b t)}} $
 $ \Rightarrow y = A{e^{(x\sqrt a + t\sqrt b )}} $
 $ A = 5{e^{ - 2}}m $
We can find the direction by finding gradient of
 $ \left( {x\sqrt a + t\sqrt b } \right) $
Let $ \phi = (x\sqrt a + t\sqrt b ) $
 $ \therefore \nabla \phi = \dfrac{{\delta (x\sqrt a + t\sqrt b )}}{{\delta x}} $ since $ [\dfrac{\delta }{{\delta x}}(Kx)] = K $
 $ \nabla \phi = \sqrt a $
 $ \hat \nabla \phi = \dfrac{{\nabla \phi }}{{\left| {\nabla \phi } \right|}} $
 $ \Rightarrow \hat \nabla \phi = \dfrac{{\sqrt a }}{{\sqrt {a + b} }}\hat i $ since $ \left| {\nabla \phi } \right| = \sqrt {{{(\sqrt a )}^2} + {{(\sqrt b )}^2}} $
here, There is positive sign between $ x $ and $ t $
Wave travels in $ ( - ) $ x direction
For Speed of wave we compare wave equation with standard wave equation having form
 $ y = A'{e^{(kx + \omega t)}} $
We found $ k = \sqrt a $ and $ \omega = \sqrt b $
So by using dispersion relation $ {\omega ^2} = {v^2} \times {k^2} $
Putting values we get $ \Rightarrow {v^2} = \dfrac{{{\omega ^2}}}{{{k^2}}} $
 $ \Rightarrow {v^2} = \dfrac{b}{a} $
 $ v = \sqrt {\dfrac{9}{{25}}} $
 $ \Rightarrow v = \dfrac{3}{5}m{s^{ - 1}} $
For maximum displacement we compare with standard wave equation $ y = A'{e^{(kx + \omega t)}} $
We found $ A = A' $
So maximum displacement is $ A = 5{e^{ - 2}} = 0.67m $
So the correct options are [B] Travelling wave is propagated along $ ( - ) $ x direction
And [C] Speed of wave is $ \dfrac{3}{5}m{\sec ^{ - 1}} $ .

Note :
It should be remembered that wave direction depends on the sign between x and t. If there is $ ( + ) $ sign between then wave direction would be opposite of direction of gradient $ \hat \nabla $ and if there is $ ( - ) $ sign between x and t then wave direction would be in direction of gradient $ \hat \nabla $ . Also dispersion relation of the electromagnetic wave is a relation between its angular frequency, wave velocity and the wave number.