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

Vedclass · Accepted Answer

$1.256$

Vedclass · Answer

(B) The standard equation for a stationary wave is given by $y = A \sin(kx) \cos(\omega t)$.Comparing this with the given equation $y = 0.15 \sin(5x) \cos(300t)$,we identify the wave number $k = 5 \, 	ext{rad/m}$.The relationship between wave number $k$ and wavelength $\lambda$ is $k = \frac{2\pi}{\lambda}$.Substituting the value of $k$: $5 = \frac{2\pi}{\lambda}$.Solving for $\lambda$: $\lambda = \frac{2\pi}{5} = 0.4 	imes 3.14159 \approx 1.2566 \, 	ext{m}$.Thus,the wavelength is approximately $1.256 \, 	ext{m}$.

For a stationary wave represented by the equation $y = 0.15 \sin(5x) \cos(300t) \, \text{m}$,what is the wavelength of the wave (in $, \text{m}$)?

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