A wave pulse is given by the equation y = f(x, t) = A exp (–B(x–vt)2). Given A =1.0m, B = 1.0 m–2 an

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A wave pulse is given by the equation y = f(x, t) = A exp (–B(x–vt)2). Given A =1.0m, B = 1.0 m–2 and v = + 2.0 m/s, which of the following graph shows the correct wave profile at the instant t = 1 s?

Accepted Answer

Sol. y = A$e^{\lbrack - B(x - {vt})^{2}\rbrack}$ A = 1, B = 1, v = 2m/s.At t = 1sec.y = $e^{\lbrack - (x - 2)^{2}\rbrack}$y = $e^{\lbrack 4x - x^{2} - 4\rbrack}$At t = 1 secy = 1 at x = 2 point of maxima

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