Respuesta :
Answer:
T = 4.42 10⁴ N
Explanation:
this is a problem of standing waves, let's start with the open tube, to calculate the wavelength
λ = 4L / n n = 1, 3, 5, ... (2n-1)
How the third resonance is excited
m = 3
L = 192 cm = 1.92 m
λ = 4 1.92 / 3
λ = 2.56 m
As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio
v = λ f
f = v / λ
f = 343 / 2.56
f = 133.98 Hz
Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m
The expression for standing waves on a string is
λ = 2L / n
λ = 2 0.37 / 2
λ = 0.37 m
The speed of the wave is
v = λ f
As we have some resonance processes between the string and the tube the frequency is the same
v = 0.37 133.98
v = 49.57 m / s
Let's use the relationship of the speed of the wave with the properties of the string
v = √ T /μ
T = v² μ
T = 49.57² 18
T = 4.42 10⁴ N
Answer:
176.9N
Explanation:
Length of wire lw 37cm=0.37m
Length of tube Lt= 192cm=1.92cm
Linear density of wire= 18*10^-3 kg/m
Speed of sound in air v = 343m/s
The third vibrational mode of tube is f3=3[v/4Lt]=3(343/4*1.92) =133.98Hertz
Recall speed of sound is ✓(Tension /linear density)
Also
F= v/2Lw
Hence
Subtituting v=✓(Tension/linear density)
We have
1/2L✓(tension/ linear density)=133.98
Tension=(133.98*2*0.37)^2*18*10^-3
Tension=176.9N
