The wave described by $y=0.25 \sin (10 \pi x-2 \pi t)$, where $x$ and $y$ are in metre and $t$ in second, is a wave travelling along the

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Aicharya · Accepted Answer

Key Idea : The sign between two terms in argument of sine will define its direction.
Writing the given wave equation
$y=0.25 \sin (10 \pi x-2 \pi t)$
The minus $(-)$ between $(10 \pi x)$ and $(2 \pi t)$ implies that the wave is travelling along positive $x$ direction.
Now comparing Eq. (i) with standard wave equation
$y=a \sin (k x-\omega t)$
We have
$\begin{array}{l}
a=0.25 \mathrm{~m}, \omega=2 \pi, k=10 \pi \mathrm{m} \\
\therefore \frac{2 \pi}{T}=2 \pi \\
\Rightarrow f=1 \mathrm{~Hz}
\end{array}$
Also, $\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{10 \pi}=0.2 \mathrm{~m}$
Therefore, the wave is travelling along +ve $x$ direction with frequency $1 \mathrm{~Hz}$ and wavelength $0.2 \mathrm{~m}$

The wave described by $y=0.25 \sin (10 \pi x-2 \pi t)$, where $x$ and $y$ are in metre and $t$ in second, is…

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