# A cord carries a wave of average power 20 mW and period 40 ms. ?

A cord carries a wave of average power 20 mW and period 40 ms. The cord has mass density

3 g/m and is tensioned to 15 N. What is the amplitude of the wave?

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- NCSLv 72 months agoFavourite answer
P = ½µω²A²v

P = 20E-3 W

µ = 3E-3 kg/m

ω = 2π / period = 2π / 4E-2s = 157 rad/s

v = √(T/ρ) = √(15N / 0.003kg/m) = 70.7 m/s

Dropping units for ease (A is in meters)

20E-3 = ½ * 0.003 * 157² * A² * 70.7

A² = 7.6e-6 m²

A = 0.0027 m = 2.7 mm

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- FiremanLv 72 months ago
BY v = √[T/µ]

=>v = √[15/(3 x 10^-3)]

=>v = √[5 x 10^-3]

=>v = 7.07 x 10^-2 m/s

As ω = 2π/T = (2 x 3.14)/(40 x 10^-3)

=>ω = 157 rad/sec

Now BY E = 1/2µω^2 A^2v

=>20 x 10^-3 = 1/2 x 3 x 10^-3 x (157)^2 x A^2 x 7.07 x 10^-2

=>A = √[7.65 x 10^-3]

=>A = 8.75 x 10^-2 m OR 8.75 cm

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