A transverse wave is propagating in stretched string along positive x-direction. The density of medium is $450 k g / m^{3}$ . The oscillation energy density of particle in the medium is $0.25 π^{2} J / m^{3}$ and maximum shear strain produced in the medium is $3 π \times 10^{- 6}$ . If the phase difference between particle A and B is $72^{\circ}$ as shown in the figure then the equation of wave is best represented by (all quantities are in S.I units)

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$(15 \times 10^{- 6}) \sin (\frac{40000}{9} π t - \frac{π}{5} x)$

$(15 \times 10^{- 6}) \sin (\frac{20000}{9} π t - \frac{2 π}{5} x)$

$(\frac{15}{2} \times 10^{- 6}) \sin (\frac{20000}{9} π t - \frac{2 π}{5} x)$

$(\frac{15}{2} \times 10^{- 6}) \sin (\frac{40000}{9} π t - \frac{2 π}{5} x)$

answer is D.

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Detailed Solution

Energy density $= \frac{1}{2} ρ A^{2} ω^{2}$
$0.25 π^{2} = \frac{1}{2} (450) A^{2} {(2 π f)}^{2} \frac{1}{60} = A f - - - (i)$
Maximum shear strain = AK
$A \frac{2 π}{λ} = 3 π \times 10^{- 6} \frac{A}{λ} = \frac{3}{2} \times 10^{- 6} - - - (i i) (72) \times \frac{π}{180} = \frac{2 π}{λ} (1) λ = 5 m S o, A = 7.5 \times 10^{- 6} m A f = 1 / 60 f = \frac{20000}{9} ω = \frac{40000 π}{9} y = (\frac{15}{2} \times 10^{- 6}) \sin (\frac{40000}{9} π t - \frac{2 π}{5} x)$

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