Question

At \[t = 0\], the shape of a travelling pulse is given by,
\[y\left( {x,0} \right) = \dfrac{{4 \times {{10}^{ - 3}}}}{{8 - {x^2}}}\] where x and y are in metres. The wave function for the travelling pulse if the velocity of propagation is 5 m/s in the x-direction is given by,
A. \[y\left( {x,t} \right) = \dfrac{{4 \times {{10}^{ - 3}}}}{{8 - \left( {{x^2} - 5t} \right)}}\]
B. \[y\left( {x,t} \right) = \dfrac{{4 \times {{10}^{ - 3}}}}{{8 - {{\left( {{x^2} - 5t} \right)}^2}}}\]
C. \[y\left( {x,t} \right) = \dfrac{{4 \times {{10}^{ - 3}}}}{{8 - {{\left( {{x^2} + 5t} \right)}^2}}}\]
D. \[y\left( {x,t} \right) = \dfrac{{4 \times {{10}^{ - 3}}}}{{8 - \left( {{x^2} + 5t} \right)}}\]

Answer 1

Hint:If the wave is travelling along –x axis then the displacement of the shape of the wave should decrease. If the wave is travelling along the +x axis then the displacement of the shape of travelling should increase.

Formula used:
\[y\left( {x,t} \right) = f\left( {x - vt} \right)\] is the representation of the displacement of the travelling pulse as a function of the time and horizontal position.

Complete step by step solution:
At \[t = 0\], the shape of a travelling pulse is given by,
\[y\left( {x,0} \right) = \dfrac{{4 \times {{10}^{ - 3}}}}{{8 - {x^2}}}\]
It is given that the wave is travelling in the x-direction. So, the rate of change of position along the x axis will be equal to the wave speed. The wave speed of the travelling wave is given as 5 m/s. So, after time t the position of the shape of the travelling pulse will be rightward and it is represented as the function of \[\left( {x - vt} \right)\].

So, the travelling pulse will be written as,
\[y\left( {x,t} \right) = f\left( {x - vt} \right)\]
Replacing x in the equation for the shape of the travelling wave at \[t = 0\]with \[\left( {x - vt} \right)\] we get the position of the shape of the travelling pulse at any time t.
\[y\left( {x,t} \right) = \dfrac{{4 \times {{10}^{ - 3}}}}{{8 - {{\left( {x - vt} \right)}^2}}}\]
Here, v is the wave speed which is equal to 5 m/s.

So, the wave function becomes,
\[y\left( {x,t} \right) = \dfrac{{4 \times {{10}^{ - 3}}}}{{8 - {{\left( {x - 5t} \right)}^2}}}\]
So, the shape of the travelling wave at any time t and at any position along horizontal x can be given as,
\[y\left( {x,t} \right) = \dfrac{{4 \times {{10}^{ - 3}}}}{{8 - {{\left( {x - 5t} \right)}^2}}}\]

Therefore, the correct option is B.

Note: If the travelling pulse is moving leftward then the displacement should decrease. In that case we should be writing \[y\left( {x,t} \right) = f\left( {x + vt} \right)\].