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(B) The standard equation for a traveling wave is given by $y(x, t) = A \cos(kx - \omega t)$,where $k = \frac{\omega}{v}$.For the first wave $y_1 = 0.

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) The standard equation for a traveling wave is given by $y(x, t) = A \cos(kx - \omega t)$,where $k = \frac{\omega}{v}$.For the first wave $y_1 = 0.05 \cos(0.50 \pi x - 100 \pi t)$:Comparing with the standard equation,we have angular frequency $\omega_1 = 100 \pi 	ext{ rad/s}$ and wave number $k_1 = 0.50 \pi 	ext{ rad/m}$.The velocity $v_1$ is given by $v_1 = \frac{\omega_1}{k_1} = \frac{100 \pi}{0.50 \pi} = 200 	ext{ m/s}$.For the second wave $y_2 = 0.05 \cos(0.46 \pi x - 92 \pi t)$:Comparing with the standard equation,we have angular frequency $\omega_2 = 92 \pi 	ext{ rad/s}$ and wave number $k_2 = 0.46 \pi 	ext{ rad/m}$.The velocity $v_2$ is given by $v_2 = \frac{\omega_2}{k_2} = \frac{92 \pi}{0.46 \pi} = 200 	ext{ m/s}$.Since both waves have the same velocity,the velocity of the waves is $200 	ext{ m/s}$.

When two sound waves travel in the same direction in a medium,the displacements of a particle located at $x$ at time $t$ are given by:
$y_1 = 0.05 \cos(0.50 \pi x - 100 \pi t)$
$y_2 = 0.05 \cos(0.46 \pi x - 92 \pi t)$
Then the velocity of the waves is..... $m/s$.

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When two sound waves travel in the same direction in a medium,the displacements of a particle located at $x$ at time $t$ are given by:$y_1 = 0.05 \cos(0.50 \pi x - 100 \pi t)$$y_2 = 0.05 \cos(0.46 \pi x - 92 \pi t)$Then the velocity of the waves is..... $m/s$.