A piano wire with mass 3.00 g and length 80.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplitude 1.6 mm travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

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Esther Phillips · Accepted Answer

Step 1KnownsnormalsizeWe know that the following equation gives the average power carried by a sinusoidal wave on a string:Pavg=12μFω2A2(1)And the following equation describes the relationship between a wave&#039;s frequency and angular speed:ω=2πf&amp;nbsp;(R.1)Step 2GivennormalsizeThe mass of the wire ism=3.00g=0.003kg, its length isl=80.0 cm=0.800 mand the tension in it is&amp;nbsp;F=25.0N&amp;nbsp;;The wave traveling along the wire has frequency of&amp;nbsp;f=120.0Hz&amp;nbsp;and its amplitude is1.6mm=1.60×10−3mStep 3Calculationsnormalsize(A) We are aware that the mass per unit length corresponds to the linear mass density; Consequently, the wire&#039;s linear mass density is given as follows:μ=ml=0.003kg0.800m=3.75×10−3kg/mWe substitute for f into relation (R.1), so we get:ω=2π(120s−1)=754s−1Now, we plug our values for&amp;nbsp;μ, F,&amp;nbsp;ω&amp;nbsp;and A into equation (1), so we get:Pavg=12(3.75×10−3)×(25.0)⋅(754)2⋅(1.60×10−3)∴Pavg=0.223W

Maria Lopez · Accepted Answer

Step 4&amp;nbsp;(b) We can observe from equation (1) that the average power is related to the amplitude&#039;s square;Pavg∝A2⇒Pavg,2Pavg,1=(A2A1)2&amp;nbsp;As a result, when the amplitude is cut in half, the average power is cut in half as well:Pavg,2Pavg,1=(A12A1)2=14&amp;nbsp;​Pavg,2=Pavg4=0.2234=0.056W&amp;nbsp;∴Pavg,2=0.056W

A piano wire with mass 3.00 g and length 80.0 cm is stretched with a t

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