The standing wave in a medium is expressed as | vedclass

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

Question

(D) The given equation of the standing wave is $y = 0.2 \sin(0.8 x) \cos(3000 t) \, m$.Comparing this with the standard equation $y = A \sin(kx) \cos(

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The given equation of the standing wave is $y = 0.2 \sin(0.8 x) \cos(3000 t) \, m$.Comparing this with the standard equation $y = A \sin(kx) \cos(\omega t)$,we get the wave number $k = 0.8 \, rad/m$.The relationship between wave number $k$ and wavelength $\lambda$ is $k = \frac{2\pi}{\lambda}$.Thus,$\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.8} = \frac{20\pi}{8} = 2.5\pi \, m$.The distance between two consecutive points of minimum displacement (nodes) or maximum displacement (antinodes) is given by $\frac{\lambda}{2}$.Distance $= \frac{\lambda}{2} = \frac{2.5\pi}{2} = 1.25\pi \, m = \frac{5\pi}{4} \, m$.Since $\frac{5\pi}{4} \, m$ is not among the given options,the correct choice is $(d)$.

The standing wave in a medium is expressed as $y = 0.2 \sin(0.8 x) \cos(3000 t) \, m$. The distance between any two consecutive points of minimum or maximum displacement is

Similar Questions

The speed of a stationary wave represented by the equation $y = 0.7 \sin \left(\frac{7 \pi}{4} x\right) \cos (350 \pi t)$ is (In the given equation $x$ and $y$ are in metre and $t$ is in second). (in $m \ s^{-1}$)

What are stationary waves? Obtain their equation.

In stationary waves,the distance between a node and its nearest antinode is $20 \ cm$. The phase difference between two particles having a separation of $60 \ cm$ will be

The wave pattern on a stretched string is shown in the figure. Interpret what kind of wave this is and find its wavelength.

In stationary waves,all particles between two consecutive nodes pass through the mean position:

Vedclass Test Series

Exam Paper Generator

Online Exam Module