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For the travelling harmonic wave y(x, t)=6 \cos 3 \pi(7 t-0.009 x+0.27) where x and y are in cm and t is in s. The phase difference between oscillatory motion of two points separated by a distance 0.6 \mathrm{~m}.

Option: 1

1.62\pi rad


Option: 2

1.82\pi rad


Option: 3

1.95\pi rad


Option: 4

1.65\pi rad


Answers (1)

y=6 \cos (21 \pi t-0.027 \pi x+0.81 \pi)

The standard equation of travelling harmonic wave is

 y=a \cos (\omega t-k n+\varphi) \\

k=0.027 \pi                             \frac{2 \pi}{\lambda}=0.027 \pi \\

Or  \lambda=\frac{2}{0.027} \mathrm{~cm}

Phase difference =\frac{2 \pi}{\lambda} \times Path difference

or  \Delta \varphi=\frac{2 \pi}{\lambda} \times \Delta x

When \Delta x=0.6 \mathrm{~m}=60 \mathrm{~cm}

\therefore \Delta \varphi=2 \pi \times \frac{0.027}{2} \times 60=1.62 \pi \mathrm{rad}

Posted by

Sumit Saini

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