The displacement of a wave is given by y=0.00 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The standard equation for a progressive wave is $y = a \sin(\omega t + kx)$.Comparing the given equation $y = 0.002 \sin(100t + x)$ with the stand

Vedclass · Accepted Answer

travelling with a velocity of $100 \ m/s$ in the negative $x$-direction

Vedclass · Answer

(B) The standard equation for a progressive wave is $y = a \sin(\omega t + kx)$.Comparing the given equation $y = 0.002 \sin(100t + x)$ with the standard form,we get:Angular frequency $\omega = 100 \ rad/s$Wave number $k = 1 \ rad/m$Amplitude $a = 0.002 \ m$Frequency $f = \frac{\omega}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi} \ Hz$.Wave velocity $v = \frac{\omega}{k} = \frac{100}{1} = 100 \ m/s$.Since the sign between $\omega t$ and $kx$ is positive,the wave is travelling in the negative $x$-direction.Therefore,the wave is travelling with a velocity of $100 \ m/s$ in the negative $x$-direction.

The displacement of a wave is given by $y=0.002 \sin (100t + x)$,where $x$ and $y$ are in meters and $t$ is in seconds. This represents a wave:

Similar Questions

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

The graph between wave number $(\bar \nu)$ and angular frequency $(\omega)$ is

$A$ simple harmonic progressive wave is given by $Y = Y_0 \sin 2 \pi (nt - \frac{x}{\lambda})$. If the wave velocity is $(1/8)^{\text{th}}$ of the maximum particle velocity,then the wavelength is

$A$ transverse wave $Y=2 \sin (0.01 x+30 t)$ moves on a stretched string from one end to another end in $0.5 \ s$. If $x$ and $y$ are in $cm$ and $t$ in $s$,then the length of the string is: (in $m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module