A progressive wave moving along the x-axis is | vedclass

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(A) The given wave equation is $y = A \sin \left[ \frac{2 \pi}{\lambda} (vt - x) \right]$.Comparing this with the standard form $y = A \sin (\omega t

Vedclass · Accepted Answer

$2 \pi A / 3$

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(A) The given wave equation is $y = A \sin \left[ \frac{2 \pi}{\lambda} (vt - x) \right]$.Comparing this with the standard form $y = A \sin (\omega t - kx)$, we have $\omega = \frac{2 \pi v}{\lambda}$ and $k = \frac{2 \pi}{\lambda}$.The particle velocity $v_p$ is given by $\frac{\partial y}{\partial t} = A \omega \cos \left[ \frac{2 \pi}{\lambda} (vt - x) \right]$.The maximum particle velocity is $(v_p)_{\max} = A \omega = A \left( \frac{2 \pi v}{\lambda} \right)$.According to the problem, $(v_p)_{\max} = 3v$.Therefore, $A \left( \frac{2 \pi v}{\lambda} \right) = 3v$.Canceling $v$ from both sides, we get $\frac{2 \pi A}{\lambda} = 3$.Solving for $\lambda$, we find $\lambda = \frac{2 \pi A}{3}$.

$A$ progressive wave moving along the $x$-axis is represented by $y = A \sin \left[ \frac{2 \pi}{\lambda} (vt - x) \right]$. The wavelength $\lambda$ at which the maximum particle velocity is $3$ times the wave velocity is:

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