The displacement of a wave is given by $y=0 \cdot 002 \sin (100 t+x)$ where ' $x$ 'and ' $y$ ' are in metre…
The displacement of a wave is given by $y=0 \cdot 002 \sin (100 t+x)$ where ' $x$ 'and ' $y$ ' are in metre and ' $t$ ' is in second. This represents a wave
of wavelength one metre
travelling with a velocity of $100 \mathrm{~m} / \mathrm{s}$ in the negative x -direction
of frequency $\left(\frac{100}{\pi}\right) \mathrm{Hz}$
travelling with a velocity of $\left(\frac{50}{\pi}\right) \mathrm{m} / \mathrm{s}$ in the positive x -direction
Solution
Comparing $\mathrm{y}=0.002 \sin (\omega \mathrm{t}+\mathrm{x})$ with $\mathrm{y}=\mathrm{a} \sin$ ( $\omega \mathrm{t}+\mathrm{kx}$ )
We get $\omega=100 \mathrm{rad} / \mathrm{s}$ and $\mathrm{k}=1 \mathrm{rad} / \mathrm{m}$ and $\mathrm{a}=0.002$
$\therefore \quad \mathrm{n}=\frac{\omega}{2 \pi}=\frac{100}{2 \pi}=\frac{50}{\pi} \mathrm{~Hz}$
Velocity $\mathrm{v}=\frac{\omega}{\mathrm{k}}=100 \mathrm{~m} / \mathrm{s}$ and $\mathrm{k}=\frac{2 \pi}{\lambda}$
$\therefore \quad \lambda=\frac{2 \pi}{\mathrm{k}}=2 \pi \mathrm{~m}$
Thus, it represents a wave travelling with a velocity of $100 \mathrm{~m} / \mathrm{s}$ in the -ve direction.