Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Waves
0 votes

The displacement $y$ of a particle in a medium can be expressed as $y = 10^{-6} \sin \bigg(100t + 20x + \large\frac{\pi}{4}\bigg)m$, where $t$ is in second and $x$ is in metre. The speed of the wave is

$\begin {array} {1 1} (a)\;2000 m/s & \quad (b)\;5 m/s \\ (c)\;20 m/s & \quad (d)\;5 \pi m/s \end {array}$


Can you answer this question?

1 Answer

0 votes
Comparing the given equation with $ y = a \sin(\omega t + kx + \phi),$
we get,$ \omega = 100\: rad/s,\: k = 20\: m^{-1}$
So, $v = \large\frac{ \omega}{k} = \large\frac{100}{20} = 5\: m/s$
Ans : (b)
answered Feb 28, 2014 by thanvigandhi_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App