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A wave travelling in the positive $x$-direction with amplitude $\mathrm{A}=0.2 \mathrm{~m}$, velocity $v=360 \mathrm{~ms}^{-1}$ and wavelength $\lambda$ $=60 \mathrm{~m}$, then the correct expression for the wave is:
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Verified Answer
The correct answer is:
$y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]$
The general equation of wave travelling in $x$-direction is given as
$$
y=A \sin \left[\frac{2 \pi}{\lambda}(V t-x)\right]
$$
Where $A=0.2 \mathrm{~m}, v=360 \mathrm{~ms}^{-1}, \lambda=$ $60 \mathrm{~m}$
$\therefore$ The equation becomes
$$
\begin{aligned}
& y=0.2 \sin \left[\frac{2 \pi}{60}(360 t-x)\right] \\
& y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]
\end{aligned}
$$
$$
y=A \sin \left[\frac{2 \pi}{\lambda}(V t-x)\right]
$$
Where $A=0.2 \mathrm{~m}, v=360 \mathrm{~ms}^{-1}, \lambda=$ $60 \mathrm{~m}$
$\therefore$ The equation becomes
$$
\begin{aligned}
& y=0.2 \sin \left[\frac{2 \pi}{60}(360 t-x)\right] \\
& y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]
\end{aligned}
$$
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