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 number of waves contained in unit length of the medium is called

$A$ transverse wave is represented by $y = \frac{10}{\pi} \sin \left( \frac{2\pi}{T}t - \frac{2\pi}{\lambda}x \right)$. For what value of the wavelength is the wave velocity twice the maximum particle velocity (in $cm$)?

$A$ transverse wave is given by $y = A \sin 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right)$. The maximum particle velocity is equal to $4$ times the wave velocity when:

The displacement $y$ (in $cm$) in case of a simple harmonic wave is given by $y=\frac{10}{\pi} \sin \left(2000 \pi t-\frac{\pi x}{17}\right)$. The period and maximum velocity of the particles in the medium will respectively be

Vedclass Test Series

Exam Paper Generator

Online Exam Module