The equation of a transverse wave travelling on a rope is given by $y = 10\sin \pi (0.01x - 2.00t)$ where $y$ and $x$ are in $cm$ and $t$ is in $seconds$. The maximum transverse speed of a particle in the rope is about .... $cm/s$.
- A$63$
- B$75$
- C$100$
- D$121$
Explore More
Similar Questions
If the equation of a transverse wave is $y = 5\sin 2\pi \left[ \frac{t}{0.04} - \frac{x}{40} \right]$,where distance is in $cm$ and time is in seconds,then the wavelength of the wave is .... $cm$.
Easy
View SolutionTwo points are located at a distance of $10\; m$ and $15 \;m$ from the source of oscillation. The period of oscillation is $0.05 \;s$ and the velocity of the wave is $300 \;m/s$. What is the phase difference between the oscillations of the two points?
MediumAIPMT 2008
View Solution$A$ progressive wave travelling along the positive $x-$ direction is represented by $y(x, t) = A \sin(kx - \omega t + \phi)$. Its snapshot at $t = 0$ is given in the figure. For this wave,the phase $\phi$ is
The phase difference between two waves represented by $y_1 = 10^{-6} \sin [100t + (x/50) + 0.5] \, m$ and $y_2 = 10^{-6} \cos [100t + (x/50)] \, m$,where $x$ is expressed in meters and $t$ is expressed in seconds,is approximately .... $rad$.
MediumAIPMT 2004
View Solution$A$ wave travelling in the positive $x-$direction with amplitude $A = 0.2\;m$ has a velocity of $v = 360\;m/s.$ If the wavelength $\lambda = 60\;m,$ then the correct expression for the wave is:
MediumAIPMT 2002
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo