For a travelling harmonic wave y(x, t) = 2.0 | vedclass

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Question

(C) The given wave equation is $y(x, t) = 2.0 \cos(2\pi(10t - 0.0080x + 0.35))$.
Comparing this with the standard wave equation $y = A \cos(2\pi ft -

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(C) The given wave equation is $y(x, t) = 2.0 \cos(2\pi(10t - 0.0080x + 0.35))$.
Comparing this with the standard wave equation $y = A \cos(2\pi ft - kx + \phi)$, we identify the wave number $k$.
The term inside the cosine function is $2\pi(10t - 0.0080x + 0.35) = 20\pi t - 0.016\pi x + 0.7\pi$.
Thus, the wave number $k = 0.016\pi 	ext{ cm}^{-1}$.
The phase difference $\Delta\phi$ between two points separated by a distance $\Delta x$ is given by $\Delta\phi = k \Delta x$.
Given $\Delta x = 0.5 	ext{ m} = 50 	ext{ cm}$.
Substituting the values, $\Delta\phi = (0.016\pi 	ext{ cm}^{-1}) 	imes 50 	ext{ cm} = 0.8\pi 	ext{ rad}$.
Therefore, the correct option is $C$.

For a travelling harmonic wave $y(x, t) = 2.0 \cos 2\pi(10 t - 0.0080 x + 0.35)$, where $x$ and $y$ are in $\text{cm}$ and $t$ in $\text{s}$. The phase difference between oscillatory motion of two points separated by a distance of $0.5 \text{ m}$ is: (in $\pi \text{ rad}$)

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