JEE MAIN - Physics (2022 - 28th July Morning Shift - No. 8)
In the wave equation
$$ y=0.5 \sin \frac{2 \pi}{\lambda}(400 \mathrm{t}-x) \,\mathrm{m} $$
the velocity of the wave will be:
200 m/s
200$$\sqrt2$$ m/s
400 m/s
400$$\sqrt2$$ m/s
คำอธิบาย
$${v_{wave}} = \left| {{{coefficient\,of\,t} \over {coefficient\,of\,x}}} \right|$$
$$ = {{400} \over 1} = 400$$ m/s