Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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# The Equation of a Wave Travelling on a String Stretched Along The X-axis is Given by - Physics

Sum

The equation of a wave travelling on a string stretched along the X-axis is given by
$y = A e {}^- \left( \frac{x}{a} + \frac{t}{T} \right)^2 .$
(a) Write the dimensions of A, a and T. (b) Find the wave speed. (c) In which direction is the wave travelling? (d) Where is the maximum of the pulse located at t = T? At t = 2 T?

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#### Solution

Given,
Equation of the wave travelling on a string stretched along the X-axis:

$y = Ae {}^\left( \frac{x}{a} + \frac{t}{T} \right) {{}^-}^2$
(a) The dimensions of A (amplitude), T (time period) and $a = \frac{\lambda}{2\pi}$ , which will have the dimensions of the wavelength, are as follows:

$\left[ A \right] = \left[ M^0 L^1 T^0 \right]$

$\left[ T \right] = \left[ M^0 L^0 T^{- 1} \right]$

$\left[ a \right] = \left[ M^0 L^1 T^0 \right]$
(b) Wave speed, $\nu = \frac{\lambda}{T} = \frac{a}{T} \left[ \lambda = a \right]$
(c) If $y = f \left( t + \frac{x}{\nu} \right)$ , then the wave travels in the negative direction; and if $y = f \left( t - \frac{x}{\nu} \right)$ ,then the wave travels in the positive direction.
Thus, we have:

"y"="Ae"^[[(x/a)+(t/T)]^(-2)]

$= A e^{- \frac{1}{T} \left[ t + \left( \frac{xT}{a} \right) \right]^2}$

$= A e^{- \frac{1}{T}\left[ t + \frac{x}{V} \right]}$

$= A e^{- f\left[ t + \frac{x}{V} \right]}$
Hence, the wave is travelling is the negative direction.
(d) Wave speed, $v = \frac{a}{t}$ Maximum pulse at T  = $\left( \frac{a}{T} \right) \times T = a$                (Along  the  negative  x - axis)
Maximum pulse at t = 2T$\left( \frac{a}{T} \times 2T \right) = 2a$             (Along  the  negative  x - axis)

Therefore, the wave is travelling in the negative x-direction.

Is there an error in this question or solution?
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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 15 Wave Motion and Waves on a String
Q 2 | Page 323
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