Answer:
a
The output power is ![P= 0.764Watt](https://tex.z-dn.net/?f=P%3D%200.764Watt)
b
The Amplitude would decrease by ![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)
Explanation:
From the question we are told that
The diameter of the steel wire is = 1.0mm![= \frac{1}{1000} = 1.0*10^{-3}m](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B1000%7D%20%20%3D%201.0%2A10%5E%7B-3%7Dm)
The raduis of this steel wire is ![r = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B1%2A10%5E%7B-3%7D%7D%7B2%7D%20%3D%200.5%2A10%5E%7B-3%7Dm)
Now from the question we can deduce that the power output is equal to the power being transmitted by wave on the wire this is mathematically represented as
![P = \frac{1}{2} \mu w^2 A^2 v ----(1)](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cmu%20w%5E2%20A%5E2%20v%20----%281%29)
Where
is the mass per unit length of the wire
This is mathematically evaluated as
![\mu = a* \rho](https://tex.z-dn.net/?f=%5Cmu%20%3D%20a%2A%20%5Crho)
Where a is the area of the the wire = ![\pi r^2 = (3.142 * 0.5*10^{-3})^2 =7.855*10^{-7}m^2](https://tex.z-dn.net/?f=%5Cpi%20r%5E2%20%3D%20%283.142%20%2A%200.5%2A10%5E%7B-3%7D%29%5E2%20%3D7.855%2A10%5E%7B-7%7Dm%5E2)
is the density of steel with a generally value of ![7850 kg/m^3](https://tex.z-dn.net/?f=7850%20kg%2Fm%5E3)
So
![\mu = 7.855*10^{-7} *7850](https://tex.z-dn.net/?f=%5Cmu%20%3D%207.855%2A10%5E%7B-7%7D%20%2A7850)
![= 6.162*10^{-3}kg/m](https://tex.z-dn.net/?f=%3D%206.162%2A10%5E%7B-3%7Dkg%2Fm)
is velocity of the wave
This is mathematically evaluated as
![w=2 \pi f](https://tex.z-dn.net/?f=w%3D2%20%5Cpi%20f)
substituting 60Hz for f
We have
![w = 2 *3.142 * 60](https://tex.z-dn.net/?f=w%20%3D%202%20%2A3.142%20%2A%2060)
![=377.04 \ rad/s](https://tex.z-dn.net/?f=%3D377.04%20%5C%20rad%2Fs)
is the amplitude with a given value of 0.50 cm ![= \frac{0.50}{100} = 0.50 *10^{-2}m](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.50%7D%7B100%7D%20%3D%200.50%20%2A10%5E%7B-2%7Dm)
v is the linear velocity of the wave
This is mathematically evaluated as
![v = \sqrt{\frac{T}{\mu} }](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%20%7D)
Where T is the tension with a given value of ![7.5N](https://tex.z-dn.net/?f=7.5N)
![v = \sqrt{\frac{7.5}{6.162*10^{-3}} }](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B7.5%7D%7B6.162%2A10%5E%7B-3%7D%7D%20%7D)
![=34.89 m/s](https://tex.z-dn.net/?f=%3D34.89%20m%2Fs)
Substituting values into equation 1
![P = 6.162*10^{-3}* 377.04^2 * (0.5*10^{-2})^2 * 34.89](https://tex.z-dn.net/?f=P%20%3D%206.162%2A10%5E%7B-3%7D%2A%20377.04%5E2%20%2A%20%280.5%2A10%5E%7B-2%7D%29%5E2%20%2A%2034.89)
![P= 0.764Watt](https://tex.z-dn.net/?f=P%3D%200.764Watt)
Since the doubling of the frequency does not affect the amplitude and from equation one the output power is
of the Amplitude, Then the Amplitude would decrease by ![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)