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  2. Stuck here, help me understand: - Oscillations and Waves - JEE Main
# Engineering

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)
P perimeter

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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