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(C) The given wave equation is $y = 6 \cos(100t - 4x)$.Comparing this with the standard wave equation $y = A \cos(\omega t - kx)$, we get:Amplitude $A

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$2.4 	imes 10^{-5}$

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(C) The given wave equation is $y = 6 \cos(100t - 4x)$.Comparing this with the standard wave equation $y = A \cos(\omega t - kx)$, we get:Amplitude $A = 6 \mu	ext{m} = 6 	imes 10^{-6} 	ext{ m}$.Angular frequency $\omega = 100 	ext{ rad/s}$.Wave number $k = 4 	ext{ m}^{-1}$.The maximum particle velocity $v_p$ is given by $v_p = A\omega$.$v_p = (6 	imes 10^{-6} 	ext{ m}) 	imes (100 	ext{ rad/s}) = 6 	imes 10^{-4} 	ext{ m/s}$.The wave velocity $v$ is given by $v = \frac{\omega}{k}$.$v = \frac{100}{4} = 25 	ext{ m/s}$.The ratio of maximum particle velocity to wave velocity is $\frac{v_p}{v} = \frac{6 	imes 10^{-4}}{25}$.$\frac{v_p}{v} = 0.24 	imes 10^{-4} = 2.4 	imes 10^{-5}$.

The equation of a progressive wave is given by $y = 6 \cos(100t - 4x)$ where $y$ is in $\mu\text{m}$, $x$ is in metre and $t$ is in second. The ratio of the maximum particle velocity to the velocity of wave is

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