Two progressive waves Y_{1} = sin 2pi(frac{t} | vedclass

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(C) The given progressive waves are $Y_{1} = \sin 2\pi(\frac{t}{0.4} - \frac{x}{4})$ and $Y_{2} = \sin 2\pi(\frac{t}{0.4} + \frac{x}{4})$.Comparing th

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$\sqrt{2} \ m$

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(C) The given progressive waves are $Y_{1} = \sin 2\pi(\frac{t}{0.4} - \frac{x}{4})$ and $Y_{2} = \sin 2\pi(\frac{t}{0.4} + \frac{x}{4})$.Comparing these with the standard wave equation $Y = A \sin 2\pi(\frac{t}{T} \pm \frac{x}{\lambda})$,we get the wavelength $\lambda = 4 \ m$ and time period $T = 0.4 \ s$.When two waves of equal amplitude and frequency traveling in opposite directions superpose,they form a standing wave given by $Y = Y_{1} + Y_{2} = 2A \cos(\frac{2\pi x}{\lambda}) \sin(\frac{2\pi t}{T})$.Here,the amplitude of the standing wave at any position $x$ is given by $A_{res} = |2A \cos(\frac{2\pi x}{\lambda})|$.Given $A = 1$,$\lambda = 4 \ m$,and $x = 0.5 \ m$,we substitute these values:$A_{res} = 2 	imes 1 	imes |\cos(\frac{2\pi 	imes 0.5}{4})|$$A_{res} = 2 \cos(\frac{\pi}{4})$Since $\cos(45^{\circ}) = \frac{1}{\sqrt{2}}$,we get $A_{res} = 2 	imes \frac{1}{\sqrt{2}} = \sqrt{2} \ m$.

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