At t = 0, a transverse wave pulse in a 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 is traveling in the positive x-direction with speed of 5m/s.

y = \frac{6}{(x + 5 t)^{2} + 3}

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y = \frac{6}{(x - 5 t)^{2} + 3}

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y = \frac{6}{(x - \frac{t}{5})^{2} + 3}

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y = \frac{6}{(x + \frac{t}{5})^{2} + 3}

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Solution

The correct option is B $y = \frac{6}{(x - 5 t)^{2} + 3}$
The wave is traveling in positive x-direction with a speed of 5 m/s.
Replacing $x \to (x - v t) = (x - 5 t)$ , we get $y = \frac{6}{(x - 5 t)^{2} + 3}$ .

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At t = 0, a transverse wave pulse in a wire is described by the function: y=6x2+3 where x and y are in meters. Write the function y=(x,t) that describes this wave if it is traveling in the positive x-direction with speed of 5m/s.

The correct option is B y=6(x−5t)2+3The wave is traveling in positive x-direction with a speed of 5 m/s. Replacing x→(x−vt)=(x−5t), we get y=6(x−5t)2+3.

At t = 0, a transverse wave pulse in a 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 is traveling in the positive x-direction with speed of 5m/s.

The correct option is B $y = \frac{6}{(x - 5 t)^{2} + 3}$
The wave is traveling in positive x-direction with a speed of 5 m/s.
Replacing $x \to (x - v t) = (x - 5 t)$ , we get $y = \frac{6}{(x - 5 t)^{2} + 3}$ .