The water level in a vertical glass tube 1.00 m long can be adjusted to any position in the tube. A tuning fork vibrating at 653 Hz is held just over the open top end of the tube, to set up a standing wave of sound in the air-filled top portion of the tube. (That air-filled top portion acts as a tube with one end closed and the other end open.) Take the speed of sound to be 343 m/s. (a) For how many different positions of the water level will sound from the fork set up resonance in the tube’s air-filled portion

Question

contestada

Respuesta :

Netta00 Netta00 · Answer 1 · 2019-10-02T13:02:25+02:00

Answer:

4

Explanation:

Given that length of tube,L=1 m

Frequency ,f=653 Hz

We know that

V= f λ

4 L = nλ

So we can say that

[tex]L_n=\dfrac{nV}{4f}[/tex]

Where n is the odd number .

Here V=343 m/s

a)

For n=1:

[tex]L_1=\dfrac{343}{4\times 653}[/tex]

[tex]L_1=0.13\ m[/tex]

For n=3:

[tex]L_3=\dfrac{3\times 343}{4\times 653}[/tex]

[tex]L_3=0.39[/tex]

For n=5:

[tex]L_5=\dfrac{5\times 343}{4\times 653}[/tex]

[tex]L_5=0.65[/tex]

For n=7:

[tex]L_7=\dfrac{7\times 343}{4\times 653}[/tex]

[tex]L_7=0.91[/tex]

So total positions will be 4.