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A wave travelling along the x -axis is described by the equation $y(x,t)=0.005\cos \left ( \alpha x-\beta t \right )$ If the wavelength and the time period of the wave are 0.08 m and 2.0 s, respectively, then $\alpha \: and\: \beta$ in appropriate units are

Option 1)
$\alpha = 12.50\pi ,\beta = \frac{\pi }{2.0}$

Option 2)
$\alpha = 25.00\pi ,\beta = \pi$

Option 3)
$\alpha = \frac{0.08}{\pi},\beta = \frac{2.0}{\pi}$

Option 4)
$\alpha = \frac{0.04}{\pi},\beta = \frac{1.0}{\pi}$

Answers (1)

best_answer

As we learnt in

Relation between phase difference and path difference -

Phase difference $\left ( \Delta \phi \right )$

$= \frac{2\pi }{\lambda }\times path \: dif\! \! ference\left ( \Delta x \right )\\\lambda =wave\; length$

-

Travelling Wave Equation -

$y=A \sin \left ( Kx-\omega t \right )$

- wherein

$K=2\pi /\lambda$

$\omega = \frac{2\pi }{T}$

$\lambda =$ wave length

$T =$ Time period of oscillation

The wave travelling along the $x-axis$ is given by

$y(x,t)=0.005\; cos(\alpha x-\beta t).$

$Therefore\; \alpha =k=\frac{2\pi }{\lambda }.As\; \lambda =0.08\, m.$

$\therefore \; \; \alpha =\frac{2\pi }{0.08}=\frac{\pi }{0.04}\Rightarrow \alpha =\frac{\pi }{4}\times 100.00=25.00\, \pi$

$\omega =\beta \Rightarrow \frac{2\pi }{2.0}=\beta \Rightarrow \pi$

$\therefore \; \; \alpha =25.00\, \pi ,\beta =\pi$

Correct option is 2.

Option 1)

$\alpha = 12.50\pi ,\beta = \frac{\pi }{2.0}$

This is an incorrect option.

Option 2)

$\alpha = 25.00\pi ,\beta = \pi$

This is the correct option.

Option 3)

$\alpha = \frac{0.08}{\pi},\beta = \frac{2.0}{\pi}$

This is an incorrect option.

Option 4)

$\alpha = \frac{0.04}{\pi},\beta = \frac{1.0}{\pi}$

This is an incorrect option.

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