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(B) The given wave equation is $y = 3 \sin(36t + 0.018x + \frac{\pi}{4})$.Comparing this with the standard wave equation $y = A \sin(\omega t + kx + \

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$3$

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(B) The given wave equation is $y = 3 \sin(36t + 0.018x + \frac{\pi}{4})$.Comparing this with the standard wave equation $y = A \sin(\omega t + kx + \phi)$,we get the wave number $k = 0.018 \, cm^{-1}$.The distance between two successive crests is equal to the wavelength $\lambda$.The relationship between wavelength and wave number is $\lambda = \frac{2\pi}{k}$.Substituting the value of $k$: $\lambda = \frac{2 	imes 3.14159}{0.018} \approx 349.06 \, cm$.Since $100 \, cm = 1 \, m$,we have $\lambda = \frac{349.06}{100} \approx 3.49 \, m$.Rounding to the nearest given option,the value is $3.5 \, m$.

$A$ transverse harmonic wave on a string is described by $y = 3 \sin(36t + 0.018x + \frac{\pi}{4})$,where $x$ and $y$ are in $cm$ and $t$ is in $s$. The least distance between two successive crests in the wave is .... $m$. (in $.5$)

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