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If a progressive wave is represented as $y=2 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ where $\mathrm{x}$ is in metre and $\mathrm{t}$ is in second, then the distance travelled by the wave in $5 \mathrm{~s}$ is
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Verified Answer
The correct answer is:
$10 \mathrm{~m}$
$\begin{array}{l}
\text { Given, } y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right) \\
=3 \sin \left(\frac{\pi t}{2}-\frac{\pi x}{4}\right)
\end{array}$
Comparing it with standard equation
$\begin{array}{l}
y=r \sin \frac{2 \pi}{\lambda}(v t-x) \\
=r \sin \left(\frac{2 \pi v t}{\lambda}-\frac{2 \pi x}{\lambda}\right)
\end{array}$
We have, $\frac{2 \pi \mathrm{v}}{\lambda}=\frac{\pi}{2}$ or $\mathrm{v}=\frac{\lambda}{4}$
and $\frac{2 \pi}{\lambda}=\frac{\pi}{4}$ or $\lambda=8 \mathrm{~m} \therefore \mathrm{v}=\frac{8}{4}=2 \mathrm{~m} / \mathrm{s}$
So, the distance travelled by wave in $t$ second $=\mathrm{vt}=2 \times 5=10 \mathrm{~m}$
\text { Given, } y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right) \\
=3 \sin \left(\frac{\pi t}{2}-\frac{\pi x}{4}\right)
\end{array}$
Comparing it with standard equation
$\begin{array}{l}
y=r \sin \frac{2 \pi}{\lambda}(v t-x) \\
=r \sin \left(\frac{2 \pi v t}{\lambda}-\frac{2 \pi x}{\lambda}\right)
\end{array}$
We have, $\frac{2 \pi \mathrm{v}}{\lambda}=\frac{\pi}{2}$ or $\mathrm{v}=\frac{\lambda}{4}$
and $\frac{2 \pi}{\lambda}=\frac{\pi}{4}$ or $\lambda=8 \mathrm{~m} \therefore \mathrm{v}=\frac{8}{4}=2 \mathrm{~m} / \mathrm{s}$
So, the distance travelled by wave in $t$ second $=\mathrm{vt}=2 \times 5=10 \mathrm{~m}$
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