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(B) The time period of the wave is $T = \frac{1}{f} = \frac{1}{256} \, s \approx 3.906 	imes 10^{-3} \, s$.The standing wave oscillates between the e

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$10^{-3}$

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(B) The time period of the wave is $T = \frac{1}{f} = \frac{1}{256} \, s \approx 3.906 	imes 10^{-3} \, s$.The standing wave oscillates between the extreme position (shown at $t=0$) and the mean position (where all particles are at zero displacement,shown at $t=?$).The time taken to go from the extreme position to the mean position is an odd multiple of a quarter of the time period,i.e.,$t = (2n+1) \frac{T}{4}$ for $n = 0, 1, 2, \dots$.Calculating the possible values:For $n=0$: $t = \frac{T}{4} = \frac{1}{256 	imes 4} = \frac{1}{1024} \approx 0.976 	imes 10^{-3} \, s \approx 9.8 	imes 10^{-4} \, s$.For $n=1$: $t = \frac{3T}{4} = 3 	imes 0.976 	imes 10^{-3} \approx 2.928 	imes 10^{-3} \, s \approx 2.9 	imes 10^{-3} \, s$.For $n=2$: $t = \frac{5T}{4} = 5 	imes 0.976 	imes 10^{-3} \approx 4.88 	imes 10^{-3} \, s \approx 4.9 	imes 10^{-3} \, s$.Comparing these with the given options,$10^{-3} \, s$ is not a possible value.

The pattern of standing waves formed on a stretched string at two instants of time (extreme,mean) are shown in the figure. The velocity of the two waves superimposing to form stationary waves is $360 \, ms^{-1}$ and their frequencies are $256 \, Hz$. Which is not a possible value of $t$ (in $\sec$)?

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