Respuesta :
Answer:
a) [tex] P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W[/tex]
b) For this case we se that [tex] P \propto A^2 f^2[/tex]
Since the power is constant but the frequency is doubled we will see that [tex] A^2 \propto \frac{P}{f^2}[/tex]
So the original amplitude is [tex] A_i \propto \sqrt{\frac{P}{f^2}}[/tex]
And if the frequency is doubled we have that:
[tex] A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}[/tex]
[tex] A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}[/tex]
So then we will see that the amplitude would be reduced the half and for this case:
[tex] A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m[/tex]
Explanation:
For this case we have the following data given:
[tex] D= 1mm = 0.001 m [/tex] represent the diameter
[tex] r = D/2= 0.0005m [/tex] represent the radius
[tex] T= 5.7 N[/tex] represent the tension
[tex] f = 57 Hz[/tex] represent the frequency of the oscillator
[tex] A= 0.54 cm = 0.0054 m[/tex] represent the amplitude of the wave
Part a
For this case we can assume that the power transmitted to the wave is the same power of the oscillator. and we have the following formula for the power:
[tex] P= 2 \pi^2 \rho S v f^3 A^2[/tex]
This expression can be written in different ways:
[tex] P= 2 \pi^2 \rho S \sqrt{\frac{T}{\mu}} f^2 A^2[/tex]
[tex] P= 2\pi^2 \rho S \sqrt{\frac{T}{\rho S}} f^2 A^2 [/tex]
[tex] P= 2 \pi^2 f^2 A^2 \sqrt{S \rho T}[/tex]
Where f is the frequency , T the tension [tex] rho= 7800 \frac{kg}{m^3}[/tex] the density of the steel, A the amplitude and [tex] S= \pi r^2[/tex] the area, so then we have everuthing in order to replace and we got:
[tex] P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W[/tex]
Part b
For this case we se that [tex] P \propto A^2 f^2[/tex]
Since the power is constant but the frequency is doubled we will see that [tex] A^2 \propto \frac{P}{f^2}[/tex]
So the original amplitude is [tex] A_i \propto \sqrt{\frac{P}{f^2}}[/tex]
And if the frequency is doubled we have that:
[tex] A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}[/tex]
[tex] A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}[/tex]
So then we will see that the amplitude would be reduced the half and for this case:
[tex] A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m[/tex]
