Wave Motion 1 Question 10

10. $Y(x, t)=\frac{0.8}{\left[(4 x+5 t)^{2}+5\right]}$ represents a moving pulse where $x$ and $y$ are in metre and $t$ is in second. Then,

(1999, 3M)

(a) pulse is moving in positive $x$-direction

(b) in $2 s$ it will travel a distance of $2.5 m$

(c) its maximum displacement is $0.16 m$

(d) it is a symmetric pulse

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Answer:

Correct Answer: 10. (b, d)

Solution:

  1. The shape of pulse at $x=0$ and $t=0$ would be as shown, in Fig. (a).

$$ y(0,0)=\frac{0.8}{5}=0.16 m $$

From the figure it is clear that $y _{\max }=0.16 m$

(a)

Pulse will be symmetric (Symmetry is checked about $y _{\max }$ ) if at $t=0$

$$ y(x)=y(-x) $$

From the given equation,

$$ \begin{aligned} y(x) & =\frac{0.8}{16 x^{2}+5} \\ y(-x) & =\frac{0.8}{16 x^{2}+5} \text { at } t=0 \\ & \text { or } \quad y(x)=y(-x) \end{aligned} $$

Therefore, pulse is symmetric.

Speed of pulse

At $t=1 s$ and $x=-1.25 m$

value of $y$ is again $0.16 m$, i.e. pulse has travelled a distance of $1.25 m$ in $1 s$ in negative $x$-direction or we can say that the speed of pulse is $1.25 m / s$ and it is travelling in negative $x$-direction. Therefore, it will travel a distance of $2.5 m$ in $2 s$.

The above statement can be better understood from Fig. (b).



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