Answer :
Answer:
a
The output power is [tex]P= 0.764Watt[/tex]
b
The Amplitude would decrease by [tex]\frac{1}{2}[/tex]
Explanation:
From the question we are told that
The diameter of the steel wire is = 1.0mm[tex]= \frac{1}{1000} = 1.0*10^{-3}m[/tex]
The raduis of this steel wire is [tex]r = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m[/tex]
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
[tex]P = \frac{1}{2} \mu w^2 A^2 v ----(1)[/tex]
Where [tex]\mu[/tex] is the mass per unit length of the wire
This is mathematically evaluated as
[tex]\mu = a* \rho[/tex]
Where a is the area of the the wire = [tex]\pi r^2 = (3.142 * 0.5*10^{-3})^2 =7.855*10^{-7}m^2[/tex]
[tex]\rho[/tex] is the density of steel with a generally value of [tex]7850 kg/m^3[/tex]
So
[tex]\mu = 7.855*10^{-7} *7850[/tex]
[tex]= 6.162*10^{-3}kg/m[/tex]
[tex]w[/tex] is velocity of the wave
This is mathematically evaluated as
[tex]w=2 \pi f[/tex]
substituting 60Hz for f
We have
[tex]w = 2 *3.142 * 60[/tex]
[tex]=377.04 \ rad/s[/tex]
[tex]A[/tex] is the amplitude with a given value of 0.50 cm [tex]= \frac{0.50}{100} = 0.50 *10^{-2}m[/tex]
v is the linear velocity of the wave
This is mathematically evaluated as
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
Where T is the tension with a given value of [tex]7.5N[/tex]
[tex]v = \sqrt{\frac{7.5}{6.162*10^{-3}} }[/tex]
[tex]=34.89 m/s[/tex]
Substituting values into equation 1
[tex]P = 6.162*10^{-3}* 377.04^2 * (0.5*10^{-2})^2 * 34.89[/tex]
[tex]P= 0.764Watt[/tex]
Since the doubling of the frequency does not affect the amplitude and from equation one the output power is [tex]\ \frac{1}{2}[/tex] of the Amplitude, Then the Amplitude would decrease by [tex]\frac{1}{2}[/tex]