Two pulses traveling on the same string are describes by the function y 1 and y 2 generated from points A and B is in meter respectively y 1- 2-x-2 t2-1- y2-2-x-4 t2-2 where x in metre and t in sec are in the distance between A - B is 1-5 m- if the position where these two pulses will meet- Shape of the pulse does not change is x - 10-1 m Find x

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Byju's Answer

Standard XII

Physics

Principle of Superposition

Two pulses tr...

Question

Two pulses traveling on the same string are describes by the function $y_{1}$ and $y_{2}$ generated from points A and B is in meter respectively $y_{1} = \frac{2}{(x - 2 t)^{2} + 1}, y_{2} = \frac{- 2}{(x - 4 t)^{2} + 2}$ where x (in metre) and t (in sec) are in the distance between A & B is 1.5m, if the position where these two pulses will meet. (Shape of the pulse does not change) is $x \times 10^{- 1} m$ Find x ___

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Solution

$Y_{1}$ is travelling along positive x direction
$Y_{2}$ is travelling along negative x direction.
Speed of wave $Y_{1}$ is = 2 m/s
Speed of wave $Y_{2}$ is = –4m/s
Suppose they meet at a distance x from A
$\frac{x}{2} = \frac{1.5 - x}{4} \Rightarrow 4 x = 3 - 2 x$
6x = 3
x = 0.5 m from A.

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Y1 is travelling along positive x direction Y2 is travelling along negative x direction. Speed of wave Y1 is = 2 m/s Speed of wave Y2 is = –4m/s Suppose they meet at a distance x from A x2=1.5−x4⇒4x=3−2x 6x = 3 x = 0.5 m from A.

$Y_{1}$ is travelling along positive x direction
$Y_{2}$ is travelling along negative x direction.
Speed of wave $Y_{1}$ is = 2 m/s
Speed of wave $Y_{2}$ is = –4m/s
Suppose they meet at a distance x from A
$\frac{x}{2} = \frac{1.5 - x}{4} \Rightarrow 4 x = 3 - 2 x$
6x = 3
x = 0.5 m from A.