A wave travelling in the +ve x-direction having displacement along y-direction as 1 m, wavelength 2\pim and frequency of

\frac{1}{\pi}Hz is represented by:

  • Option 1)

    y=\sin(2\pi x+2\pi \text{t})

  • Option 2)

    y=\sin(x-2\text{t})

  • Option 3)

    y=\sin(2\pi x-2\pi \text{t})

  • Option 4)

    y=\sin(10\pi x-20\pi \text{t})

 

Answers (1)

As we learnt in

Relation between phase difference and path difference -

Phase difference left ( Delta phi 
ight )

= frac{2pi }{lambda }	imes path : dif! ! ferenceleft ( Delta x 
ight )\lambda =wave; length

-

 

 

 

 

y= A \sin (\omega t - k x +\phi)

Given, \:\:\lambda=2 \pi, \:\:\:f=\frac{1}{\pi}, \:\:\:4=1\omega t

k=\frac{2 \pi}{\lambda}= \frac{2 \pi}{2 \pi}= k =1

\therefore y= \sin (2t-x+\phi)

Correct option is 2.

 


Option 1)

y=\sin(2\pi x+2\pi \text{t})

Incorrect

Option 2)

y=\sin(x-2\text{t})

Correct

Option 3)

y=\sin(2\pi x-2\pi \text{t})

Incorrect

Option 4)

y=\sin(10\pi x-20\pi \text{t})

Incorrect

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