The transverse displacement of a string is given by $y (x, t) = 0.06 \sin (\frac{2 π}{3} X) \cos (120 π t)$ where x and y are in m and t in s. The length of the string is 1.5m and its mass is $3.0 \times 10^{- 2} kg$ Determine the tension in the string.

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864 N

324 N

972 N

648 N

answer is D.

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Detailed Solution

$y = 0.0 6 sin (\frac{2 π}{3} x) cos (120 π t)$

$y = 2A sin (kx) cos (ω t)$

$k = \frac{2 π}{3} ω = 120 π$

$V = \frac{ω}{k} = \frac{120 π}{2 π} \times 3 = 180 m / s$

$V = \sqrt{\frac{T}{π}} μ = \frac{3 \times 1 0^{- 2}}{1.5} = \frac{3}{150} = \frac{1}{50}$

$180 = \sqrt{\frac{T}{\frac{1}{50}}} \Rightarrow T = \frac{180 \times 180}{50} = 648 N$

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