At t = 0, the shape of a travelling pulse is given by $y (x, 0) = \frac{4 \times 10^{- 3}}{8 - (x)^{2}}$ 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

$y (x, t) = \frac{4 \times 10^{- 3}}{8 - (x - 5 t)}$

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$y (x, t) = \frac{4 \times 10^{- 3}}{8 - (x - 5 t)^{2}}$

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$y (x, t) = \frac{4 \times 10^{- 3}}{8 - (x + 5 t)^{2}}$

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$y (x, t) = \frac{4 \times 10^{- 3}}{8 - (x + 5 t)}$

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Solution

The correct option is B
$y (x, t) = \frac{4 \times 10^{- 3}}{8 - (x - 5 t)^{2}}$

$y (x, t) = f (x - v t) y = (x, 0) = \frac{4 \times 10^{- 3}}{8 - (x)^{2}}$

For a travelling wave in the +x – direction

$y (x, t) = \frac{4 \times 10^{- 3}}{8 - (x - 5 t)^{2}}$

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At t = 0, the shape of a travelling pulse is given by $y (x, 0) = \frac{4 \times 10^{- 3}}{8 - (x)^{2}}$ 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

Q. If at t = 0, a travelling wave pulse on a string is described by the function,

y = \frac{10}{(x^{2} + 2)}

. Here, x and y are in metre and t in second. The wave function representing the pulse at time t, if the pulse is propagating along positive x-axis with speed 2 m/s is

y = \frac{10}{(x - 2 t)^{2} + p}

. What is the value of p?

Q. A travelling wave pulse is given by.

y = \frac{10}{5 + (x + 2 t)^{2}}

Here, x and y are in metre and t in second. In which direction and with what velocity is the pulse propagating ? What is the amplitude of pulse?

Q. At t = 0, a transverse wave pulse in wire is described by the function;

y = \frac{6}{x^{2} + 3}

Where x and y are in meters. Write the function y(x,t) that describes this wave if it travelling in the positive x-direction with a speed of 4.50 m/s.

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At t = 0, the shape of a travelling pulse is given by y(x,0)=4×10−38−(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

The correct option is B y(x,t)=4×10−38−(x−5t)2 y(x,t)=f(x−vt)y=(x,0)=4×10−38−(x)2 For a travelling wave in the +x – direction y(x,t)=4×10−38−(x−5t)2

At t = 0, the shape of a travelling pulse is given by $y (x, 0) = \frac{4 \times 10^{- 3}}{8 - (x)^{2}}$ 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

The correct option is B
$y (x, t) = \frac{4 \times 10^{- 3}}{8 - (x - 5 t)^{2}}$

$y (x, t) = f (x - v t) y = (x, 0) = \frac{4 \times 10^{- 3}}{8 - (x)^{2}}$

For a travelling wave in the +x – direction

$y (x, t) = \frac{4 \times 10^{- 3}}{8 - (x - 5 t)^{2}}$