The equation y = 0.15 sin(5x) cos(300t) descr | vedclass

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(B) The standard equation for a stationary wave is given by $y = A \sin(kx) \cos(\omega t)$.Comparing the given equation $y = 0.15 \sin(5x) \cos(300t)

Vedclass · Accepted Answer

$1.26$

Vedclass · Answer

(B) The standard equation for a stationary wave is given by $y = A \sin(kx) \cos(\omega t)$.Comparing the given equation $y = 0.15 \sin(5x) \cos(300t)$ with the standard form,we identify the wave number $k = 5 	ext{ rad/m}$.The relationship between the wave number $k$ and the wavelength $\lambda$ is $k = \frac{2\pi}{\lambda}$.Substituting the value of $k$,we get $5 = \frac{2\pi}{\lambda}$.Therefore,$\lambda = \frac{2\pi}{5} = \frac{2 	imes 3.14159}{5} = \frac{6.283}{5} = 1.2566 	ext{ m}$.Rounding to two decimal places,we get $\lambda \approx 1.26 	ext{ m}$.

The equation $y = 0.15 \sin(5x) \cos(300t)$ describes a stationary wave. The wavelength of the stationary wave is .... $m$.

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