$A$ transverse progressive wave on a stretched string has a velocity of $10\,m/s$ and a frequency of $100\,Hz$. The phase difference between two particles of the string which are $2.5\,cm$ apart will be:
- A$\frac{\pi}{8}$
- B$\frac{\pi}{4}$
- C$\frac{3\pi}{8}$
- D$\frac{\pi}{2}$
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The displacement equations of two waves are given as ${y_1} = 10\sin \left( {3\pi t + \frac{\pi }{3}} \right)$ and ${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$. What is the ratio of their amplitudes?
MediumAIIMS 1997
View SolutionTwo waves are represented by the equations $y_1 = a \sin \omega t$ and $y_2 = a \cos \omega t$. The first wave
Easy
View SolutionWhen two sound waves travel in the same direction in a medium,the displacements of a particle located at $x$ at time $t$ are given by:
$y_1 = 0.05 \cos(0.50 \pi x - 100 \pi t)$
$y_2 = 0.05 \cos(0.46 \pi x - 92 \pi t)$
Then the velocity of the waves is..... $m/s$.
$y_1 = 0.05 \cos(0.50 \pi x - 100 \pi t)$
$y_2 = 0.05 \cos(0.46 \pi x - 92 \pi t)$
Then the velocity of the waves is..... $m/s$.
$A$ wave travelling along a string is described by the equation $y = A \sin (\omega t - k x)$. The maximum particle velocity is:
The equations of two simple harmonic waves are given by $Y_1 = 2 \sin 8 \pi \left(\frac{t}{0.2} - \frac{x}{2}\right) \text{ m}$ and $Y_2 = 4 \sin 8 \pi \left(\frac{t}{0.16} - \frac{x}{1.6}\right) \text{ m}$. Then both waves have:
DifficultMHT CET 2021
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