Question

A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a tube of length 192 cm that is open near the wire and closed at the other end. Please find the tension in the wire assuming that the speed of sound in air is 343 m/s.

Answer 1

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 2

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