The power in transverse wave is given as follows (see https://opentextbc.ca/universityphysicsv1openstax/chapter/16-4-energy-and-power-of-a-wave/):
P = 1 2 μ A 2 ω 2 v P = \dfrac12\mu A^2\omega^2v P = 2 1 μ A 2 ω 2 v where μ = 0.10 k g / m \mu = 0.10kg/m μ = 0.10 k g / m , A = 2.0 c m = 0.020 m A = 2.0cm = 0.020m A = 2.0 c m = 0.020 m ,
ω = 2 π f \omega = 2\pi f ω = 2 π f where f = 120 H z f = 120Hz f = 120 Hz and
v = F T μ v = \sqrt{\dfrac{F_T}{\mu}} v = μ F T where F T = 135 N F_T = 135N F T = 135 N . Combining it all together, obtain:
P = 1 2 μ A 2 ( 2 π f ) 2 F T μ = 2 π 2 μ A 2 f 2 F T μ P = 2 π 2 ⋅ 0.1 ⋅ 0.0 2 2 ⋅ 12 0 2 ⋅ 135 0.1 ≈ 4.2 × 1 0 2 W P = \dfrac12\mu A^2(2\pi f)^2\sqrt{\dfrac{F_T}{\mu}} = 2\pi^2\mu A^2f^2\sqrt{\dfrac{F_T}{\mu}}\\
P = 2\pi^2\cdot 0.1\cdot 0.02^2\cdot 120^2\cdot \sqrt{\dfrac{135}{0.1}} \approx 4.2\times 10^2W P = 2 1 μ A 2 ( 2 π f ) 2 μ F T = 2 π 2 μ A 2 f 2 μ F T P = 2 π 2 ⋅ 0.1 ⋅ 0.0 2 2 ⋅ 12 0 2 ⋅ 0.1 135 ≈ 4.2 × 1 0 2 W Answer. 4.2 × 1 0 2 W 4.2\times 10^2W 4.2 × 1 0 2 W
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