Experiment 10: Standing Waves and the Speed of Sound in Air

Object: To measure the speed of sound waves propagating in air using a resonance tube and compare the measured speed with the predicted value given by treating the air as an ideal gas.

Introduction: When a surface [ like the end of a tuning fork, or the speaker(s) in your smartphone, or your vocal cords] is vibrating , the air around it is set into vibration producing a sound wave . If the air is constrained to vibrate in a long narrow region, for example the air in a tube, Fig. 1, we have a one-dimensional wave. If we limit our analysis to sinusoidal vibrations (Harmonic Oscillator!!) then the displacement from equilibrium position, s1, at time t for the air particles having position x can be given by:

            s1  = A sin (kx    t)                ( 1 )

where  A  represents the amplitude of the oscillations for the air particles, k is the wave number, k = 2 / , with   known as the wavelength (distance between two peaks of the sinusoidal shape in space). Here kx and t are in radians, rad,  ( 2 rad = 360 ), is the angular frequency, =  2 f ,with f the frequency; the negative sign in the argument of the sine indicates that the wave propagates in the + x direction; an additional assumption has been made: the particles at  x = 0, have s1 = 0 .
The air particles are extremely small and we do not observe their motion but can  hear them when their amplitude and frequency are within the range of human hearing.
When the air in the tube is blocked at some point along its path by a material of much higher density, like a hard surface (plastic cap, solid cylinder, water ), the wave reflects back at the air-front of cap interface, giving rise to a second wave propagating along the  x  axis and having a displacement s2 ,

                s2  = A sin (kx  +  t)                ( 2 )

We are assuming that the reflected wave has the same amplitude, frequency, and wavelength as the incident wave. The net displacement of the air particles due to these two waves traveling in opposite direction is given by the sum of the displacements, eqs (1) and (2). Using some trigonometry, we find,

              s  = s1  +  s2    =  2 A sin(kx) cos(t)              ( 3 )

Question 1.Use trigonometric identities to show that, for s1 and s2 given by eqs. (1) and (2), their sum is given by eq (3) .

For particles located at points where kx is a multiple of  , the amplitude of vibration will be zero at all times, such point is referred to as a node. Adjacent nodes are separated by a distance of half a wavelength. For particles located at points where kx equals /2 , 3/2  , 5/2  , etc, the amplitude of vibration will be the largest possible, 2A; these points are referred to as antinodes. Again, the separation between adjacent antinodes, or adjacent nodes is half a wavelength.
From all this we see that the separation between a node and its nearest antinode is a quarter of a wavelength. Since the top of the plastic cap has a much higher density that air, the interface air/cap must be a node. If the mouth of the tube is made to correspond to an antinode, a louder sound will be heard; if the mouth does not correspond to an antinode, a sound will still be heard but it will not be as loud. All these can be accomplished by changing the frequency of the vibrating source or by changing the length L of the air column in Fig. 1. This is simple achieved by changing the position of the front of the cylinder, which for us will be the top of a bottle cap; the rod to hold the cup should be at least nearly as long as the tube.

   
    Fig 1. Apparatus to produce standing waves; not drawn to scale. Horizontal flat surface not shown for clarity. The signal generator is your cell phone. You could use thin book and support phone to place speaker at center of tube; also place holders at the side of tube, to avoid its rolling sideways. Calibrated scale in cm (accuracy up to 0.1 cm) not shown for clarity

Procedure:

1. Make a plastic tube with diameter about 1 mm bigger than the diameter of the plastic cap, so that the cap fits snuggly along tube; the tube should be about 20 cm long (You can do this from a plastic bottle of water with cylindrical side or a sheet of plastic from an inexpensive plastic folder, see video here). Using a ruler or graph paper in cm (smallest reading in mm) make a scale and tape it to the side of the tube, that will facilitate reading the distance, L ,  from the mouth of the tube to the surface of the cap, Fig 1. [Caution: When cutting plastic, the edges may be very sharp, make sure to cover the edges with scotch tape to smooth the edges]

2.  You will use your smartphone as the tone generator. Open your favorite online tone generator. The loudspeakers are usually at the bottom end of the phone. If there are two of them, like in iPhone, cover one of them well so very little sound comes through it. Set the audio generator to sinusoidal signal and a frequency of 2,000 Hz, enter this value on your data sheet, set volume so that the sound is faint and there is no other source of sound nearby.

3. Place loudspeaker about 1 cm above center of top of tube; you will hear the sound emitted by the loudspeaker of course. Starting with a small L value, say L 1.0 cm , move the cap to make L larger. As you slowly move the cap, there will be a position at which the intensity of the sound clearly increases, and, as you pass this level it decreases. Go back to the L value to confirm its location with accuracy of 0.1 cm (mm), called L1 and enter on your data sheet, this is your first resonance level.
At some points, L1 , L2, L3 , , also known as the resonance points of the tube, you should be able to hear a clearly louder sound: they correspond to the mouth being an antinode while the start of the solid cylinder is a node. Fig 2 indicates the amplitude of the oscillations along the air in the tube. Enter location of the resonance points with 0.1 cm precision.

                       
Fig 2. The length of the air column is controlled by the location of the air/cap interface. For resonance to occur, the mouth must correspond to a point with maximum amplitude of oscillation, an antinode, while the cap/air interface must be a node. The first four possible resonance shapes, resonance modes, are shown. For clarity, the vibrating loudspeaker at tubes mouth is not shown.

4. From Fig. 2,  L2 L1 = / 2 ; so =  2 ( L2 L1 ); similarly,  =  2 ( L3 L2 ) ; find the average wavelength 1  .

5. Determine the speed of wave propagation v1  = f1 1 .

6. Repeat steps 2 to 5 for a frequency of 2,500 Hz . From the two values for the speed of sound find the average value for the speed of sound in air, vavg

7. Measure the temperature of the air near the mouth of the tube, TC, in C, and calculate the speed of sound in air, v (If you cannot find the temperature in the room and it is comfortable in the room use 22 C )

            v    =  331  [ 1 + (TC / 273 ) ]  m/s          ( 4 )

8. Determine the percent discrepancy between the values in steps 6 and 7.

Question 2. A student in the outdoors, is 500 m from a large vertical mountain. She makes a short duration loud sound. If it takes 2.86 seconds for her to hear the echo of her loud sound, does she need a heavy jacket to be comfortable in the outdoors? Explain.

                    Data Table

First frequency            f1  =    2,000    Hz

Resonance levels        L1  =          m    ;    L2 =       m  ;    L3 =        m  ;   

Wavelength      =  2(L2  –  L1 ) =                m ;    =  2(L3  –  L2 ) =               m ; 

Average wavelength        1 =             m     

Speed of wave propagation    v1 =  1    f1  =                m/s   

Second frequency             f2  =    2,500    Hz

Resonance levels        L1  =              m    ;    L2 =             m  ;    L3 =              m  ;   

Wavelength    = 2(L2  –  L1 )=                 m ;      =  2(L3  –  L2 ) =             m ; 

Average wavelength        2 =             m     

Speed of wave propagation    v2 =     2 f2    =          m/s

Average speed of sound wave propagation    vavg  =          m/s

Temperature of region near the mouth of the tube    TC  = 

Speed of sound, from eq. ( 4 )    v =        m/s

Percent discrepancy for the two speeds, v and vavg  =