60.7k views
2 votes
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

User Nefron
by
7.1k points

1 Answer

1 vote

Answer:

4

Step-by-step 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


L_n=(nV)/(4f)

Where n is the odd number .

Here V=343 m/s

a)

For n=1:


L_1=(343)/(4* 653)


L_1=0.13\ m

For n=3:


L_3=(3* 343)/(4* 653)


L_3=0.39

For n=5:


L_5=(5* 343)/(4* 653)


L_5=0.65

For n=7:


L_7=(7* 343)/(4* 653)


L_7=0.91

So total positions will be 4.

User Beth
by
7.3k points