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A transverse wave propagating along $x$-axis is represented by $y(x, t)=8.0$ $\sin (0.5 \pi x-4 \pi t-\pi / 4)$ where $x$ is in metres and $t$ is in seconds. The speed of the wave is:
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Verified Answer
The correct answer is:
$8 \mathrm{~m} / \mathrm{s}$
It is given that:
$$
y\left(x_1 t\right)=8.0 \sin \left(0.5 \pi x-4 \pi t-\frac{\lambda}{9}\right)
$$
Compare this equation with standard equation
$$
y=a \sin \left(\frac{2 \pi x}{\lambda}-\frac{2 \pi t}{\mathrm{~T}}+\phi\right)
$$
we have
$$
\begin{aligned}
\frac{2 \pi}{\lambda} & =0.5 \pi \\
\Rightarrow \quad \lambda & =\frac{2 \pi}{0.5 \pi}=4 \mathrm{~m} \\
\frac{2 \pi}{T} & =4 \pi \\
\Rightarrow \quad T & =\frac{1}{2} \sec \\
U=\frac{1}{T} & =\frac{1}{y_2}=2 \mathrm{H}_2 .
\end{aligned}
$$
$$
y\left(x_1 t\right)=8.0 \sin \left(0.5 \pi x-4 \pi t-\frac{\lambda}{9}\right)
$$
Compare this equation with standard equation
$$
y=a \sin \left(\frac{2 \pi x}{\lambda}-\frac{2 \pi t}{\mathrm{~T}}+\phi\right)
$$
we have
$$
\begin{aligned}
\frac{2 \pi}{\lambda} & =0.5 \pi \\
\Rightarrow \quad \lambda & =\frac{2 \pi}{0.5 \pi}=4 \mathrm{~m} \\
\frac{2 \pi}{T} & =4 \pi \\
\Rightarrow \quad T & =\frac{1}{2} \sec \\
U=\frac{1}{T} & =\frac{1}{y_2}=2 \mathrm{H}_2 .
\end{aligned}
$$
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