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  2. Help me solve this - Oscillations and Waves - JEE Main-2
# Engineering

The displacement y of a particle in a medium can be expressed as:

y=10^{-6}\sin \left ( 100t +20x+\pi /4 \right )m where t is in second and x in meter. The speed of the wave is

  • Option 1)

    2000 m/s

  • Option 2)

    5 m/s

  • Option 3)

    20 m/s

  • Option 4)

    5\pi m/s

 

Answers (1)
P perimeter

As we learnt in

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

 

 

 

Given wave equation

y= 10^{-6}\sin \left ( 100t+20x+\frac{\pi }{4} \right )m

Standard equation :y= a sin\left ( \omega t+kx+\phi \right )

Compare\: the\: two

\therefore \omega = 100\: and\: k= 20

\therefore \frac{\omega }{k}= \frac{100}{20}\Rightarrow \frac{2\pi n}{2\pi /\lambda }= n\lambda = \nu = 5

\therefore \nu = 5m/s

Correct option is 2.

 


Option 1)

2000 m/s

This is an incorrect option.

Option 2)

5 m/s

This is the correct option.

Option 3)

20 m/s

This is an incorrect option.

Option 4)

5\pi m/s

This is an incorrect option.

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