Question

1.) An open vertical tube has water in it. A
tuning fork vibrates over its mouth. As the
water level is lowered in the tube, the sixth resonance is heard when the water level is
225.5 cm below the top of the tube.
What is the wavelength of the sound wave?
The speed of sound in air is 343 m/s.
Answer in units of cm.

2.)What is the frequency of the sound wave; i.e.,
the tuning fork?
Answer in units of s−1.

3.)The water continues to leak out the bottom
of the tube.
When the tube next resonates with the tuning fork, what is the length of the air column?
Answer in units of cm.

Answer 1

1) There are six nodes (N = 6) in the air column. The number of quarter wavelengths in the length of the pipe is $J=2N-1=2(6)-1=11$.

$$J=\frac{l}{\frac{\lambda}{4}}\to 11=\frac{l}{\frac{\lambda}{4}}$$

$$\lambda=\frac{4l}{11}$$

$$\lambda=\frac{4}{11}225.5=82\ cm$$

2)

$$f=\frac{v}{\lambda}=\frac{343}{0.82}=418.293\ s^{-1}$$

3) The next resonance will occur when the open vertical tube has a length of one half wavelength greater than its initial wave length.

$$l'=l+0.5\lambda=225.5+0.5(82)=266.5\ cm$$