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(D) The given equation for the stationary wave is $y = 5 \sin \left( \frac{2\pi x}{3} \right) \cos (20\pi t)$.Comparing this with the standard equatio

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$1.5$

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(D) The given equation for the stationary wave is $y = 5 \sin \left( \frac{2\pi x}{3} \right) \cos (20\pi t)$.Comparing this with the standard equation of a stationary wave $y = A \sin \left( \frac{2\pi x}{\lambda} \right) \cos (\omega t)$,we get $\frac{2\pi}{\lambda} = \frac{2\pi}{3}$.Therefore,the wavelength $\lambda = 3 \ cm$.The distance between two adjacent nodes in a stationary wave is given by $\frac{\lambda}{2}$.Distance $= \frac{3}{2} = 1.5 \ cm$.

$A$ string vibrates according to the equation $y = 5 \sin \left( \frac{2\pi x}{3} \right) \cos (20\pi t)$,where $x$ and $y$ are in $cm$ and $t$ is in $sec$. The distance between two adjacent nodes is .... $cm$.

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