A transverse wave of amplitude 0.5, m,wavelen | vedclass

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(B) The general equation for a wave traveling in the negative $x$-direction is given by $y(x, t) = A \sin(kx + \omega t + \phi)$ or $A \cos(kx + \omeg

Vedclass · Accepted Answer

$y(x, t) = 0.5\cos(2\pi x + 4\pi t)$

Vedclass · Answer

(B) The general equation for a wave traveling in the negative $x$-direction is given by $y(x, t) = A \sin(kx + \omega t + \phi)$ or $A \cos(kx + \omega t + \phi)$.Given: Amplitude $A = 0.5\, m$,Wavelength $\lambda = 1\, m$,Frequency $f = 2\, Hz$.Calculate the angular wave number $k = \frac{2\pi}{\lambda} = \frac{2\pi}{1} = 2\pi\, rad/m$.Calculate the angular frequency $\omega = 2\pi f = 2\pi(2) = 4\pi\, rad/s$.Substituting these values into the wave equation: $y(x, t) = 0.5 \cos(2\pi x + 4\pi t)$.Thus,option $B$ is correct.

$A$ transverse wave of amplitude $0.5\, m$,wavelength $1\, m$,and frequency $2\, Hz$ is propagating in a string in the negative $x$-direction. The expression for this wave is:

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