Question

Thank you and I appreciate your help and time in advance!

L = N (l/4), N = 1, 3, 5 …                                                                 (1)

Where L is the length of the air column and N is an odd integer.

When N = 1, the column has a length one quarter of wavelength with a node at the water surface and an antinode at the opening. Then the wavelength

L₁= (1/4) l, or l= 4L₁.                                                                       (2)

When N = 3, with a half-wavelength wave segment added, the air column has a length of three quarters of wavelengths with an additional node and an additional antinodes. Then the wavelength

L₂= (3/4) l, or   l= (4/3) L_2. (3)

In general, for the nth case, the string shows N wave segments. The wavelength

L _N = [(2N-1)/4] l, or l= [4/(2N-1)]L_N. (4)

It seems one might just measure the length of air column, and multiply it by 4/(2N-1) to get the wavelength. However, the position of the first antinode is not exactly at the opening, instead, it is just a little bit outside of the opening. However, we may subtract L₁ from L₂, or subtract L₁ from L₃, and etc. to eliminate the opening effect.

L₂ – L₁ = l/2,                                                                          (5)

L₃ – L₁ = l,                                                                             (6)

L₄ - L₁ = 1.5 l (7)

......

In general,       L_N+1 – L_N =l/2                                                                        (8)

and                  L_N+i – L_N = i (l/2).                                                                  (9)

For all harmonic waves, the wave speed, frequency, and wavelength are related,

                        V = fl.                                                                                    (10)

In our case, the frequency of the sound wave is the frequency of the external excitation, the vibrating tuning fork. The speed of sound is readily to be calculated.

The speed of sound depends on the temperature of the medium, in our case, air.

            V = 3311+T273                                                                                (11)

Where V is in m/s, and 331 m/s is the speed of sound in air at 0^oC, and T is the air temperature in degrees Celsius. Following this equation, one finds that at 20oC, the speed of sound is approximately 343 m/s.

QUESTIONS

1. Examine the % differences between the experiment values and theoretical value of speed of sound. By using which frequency do you reach smaller % difference with the theoretical value? Explain why.

2. Which calculations show us the precision of the measurement of the speed of sound?   By using which frequency do you reach a better precision? Explain why.

3. Do you see any possible systematic errors? Explain.

4. In Data Tables 1-2, predict the lengths of the air column for the fourth and fifth modes. Show your work.

5. Sketch a figure showing all the nodes and antinodes of the fourth mode in the resonance tube.

Answer 1

with higher frequency the measured value is closer to the theoretical value. and also more precise value. i.e the error in measurement is lesser ( the standard deviation. )

The reasons

with higher frequency the wave length is smaller and hence the part of the wave outside the tube is smaller and hence the error in measuring the wave length.

the length of the air column increases by l/2 half wave length for the next mode

Frequency 1024 Hz

wave length = 0.3334 m measured

air column for the 3rd mode = 0.521 m

4th mode = 0.521+ 0.167 = 0.688 m

5th mode = 0.688 + 0.167 = 0.855 m

Frequency 512 Hz

wave length = 0.666 m

air column for the 3rd mode = 1.189 m

4th mode = 1.189 + 0.333 = 1.522 m

5th mode = 1.522 + 0.333 = 1.855 m