Experiment 4: Standing Waves

Purpose: The objective of this lab is to test various frequencies of a standing wave on a string propelled by a forced oscillator and become familiar with the characteristics of the wave such as wavelength, frequency, wave speed, and amplitude.

Materials:

• frequency wave driver
• function generator
• string
• hanging mass
• pulley
• table/ string clamps
• meter stick for measurement

Set Up:

Procedure:

The frequency generator was responsible for changing the number of nodes on the standing wave. As the frequency went up, the number of nodes in the string increased. As the frequency went up the wavelength decreased due to the inversely proportional relationship. Wavelength can be found by this equation

which is just twice the length divide by the number of nodes in the standing wave. A distinct number of nodes can be found by setting the harmonic oscillator to a resonant frequency. For case 1 the first harmonic was at 31Hz. As the number of nodes increases, the frequency and inverse wavelength doubles, and the amplitude and wavelength are cut in half. By adjusting the frequency generator we can make the string increase its number of nodes. This table shows the number of nodes along with its wave characteristics.

Data/ Data Analysis:

1.       For cases 1, and 2, and for each frequency find the wavelength l of the wave, and the value of n that describes the wave (see equation 2).

Case 1
Nodes	Frequency (Hz)	Length (m)	Amplitude	Wavelength	1/λ
1	31	1.326	4.500	2.652	0.377
2	61	0.663	2.250	1.326	0.754
3	92	0.442	1.500	0.884	1.131
4	122	0.332	1.125	0.663	1.508
5	153	0.265	0.900	0.530	1.885
6	184	0.221	0.750	0.442	2.262
7	215	0.189	0.643	0.379	2.640
8	245	0.166	0.563	0.332	3.017
9	276	0.147	0.500	0.295	3.394
10	307	0.133	0.450	0.265	3.771

For case 2 the hanging mass was reduced to 1/4th of its original value which reduced the frequency by half, the amplitude by 1/4th, but kept the wavelength the same.

Case 2
Nodes	Frequency (Hz)	Length (cm)	Amplitude	Wavelength	1/λ
1	15	1.326	1.100	2.652	0.377
2	31	0.663	0.550	1.326	0.754
3	46	0.442	0.367	0.884	1.131
4	62	0.332	0.275	0.663	1.508
5	77	0.265	0.220	0.530	1.885
6	93	0.221	0.183	0.442	2.262

2.      For case 1, plot the frequency of the wave versus 1/l; the slope of the line should be equal to the wave speed. Compare the wave speed determined from the slope of the line with that obtained using equation 5. 

3.      Repeat step 2 for case 2, and find the slope of the line.

The slope of the line of these two graphs is the wave speed of the standing wave.
The equation for the velocity of the wave is given by:

where T is the tension in the string and μ is the linear density of the string.
The values of obtained by the slopes of the graph were

	Slope Wave Velocity (m/s)
Case 1	81.392
Case 2	41.22

The calculated velocity values obtained from the equation are

	Mass of String (g)	μ	Tension (N)	Wave Velocity
Case 1	0.74	0.000308	1.96	79.77240352
Case 2	0.74	0.000308	0.49	39.88620176

which is fairly close to our experimental values.

4.      What is the ratios of wave speeds for case 1 compared to 2 (from the graphs). Is this ratios equal to the ratio of the predicted wave speeds (Apply equation 5 here)?

In order to determine a relationship, the ratios between the two velocities of the two cases will be examined.

	Slope Wave Velocity (m/s)	Calculated Wave Velocity	Ratio 1	Ratio 2
Case 1	81.392	79.77240352	1.97	2.00
Case 2	41.22	39.88620176

The ratios are almost identical however due to the sources of error the numbers are a little off.

	n	Case 1	Case 2	Ratio
	1	31	15	2.1
	2	61	31	2.0
	3	92	46	2.0
	4	122	62	2.0
	5	153	77	2.0
	6	184	93	2.0

The ratio of the predicted wave speeds is 2.0 so the ratios are all very close to one another.

5.       Are the measured frequencies for case 1 equal to nf₁, where n is the number of the harmonic?  
f₁= v*(1/2L)

Nodes (n)	Frequency (Hz)	Length ( in meters)	nf1
1	31	1.326	30.7
2	61	0.663	61.4
3	92	0.442	92.1
4	122	0.332	122.8
5	153	0.265	153.5
6	184	0.221	184.1
7	215	0.189	214.8
8	245	0.166	245.5
9	276	0.147	276.2
10	307	0.133	306.9

The calculated values of nf₁ are very close to the values on the frequency generator.

6.      What is the ratio of the frequency of the second harmonic for case 1 compared to case 2? Is this ratio the same for the third harmonic? the fourth? the fifth? Is there a pattern here?

Nodes (n)	f1	f2	f1/f2
1	30.7	15.5	1.97
2	61.4	31.1	1.97
3	92.1	46.6	1.97
4	122.8	62.2	1.97
5	153.5	77.7	1.97
6	184.1	93.3	1.97

They are all the same ratios.

Conclusion:

This experimental values were very close to the values predicted by the equation which barely ever happens. The ratios of the harmonics were equal.