lab 10 resonant vibrations of a wire - andrews …...lab 10 resonant vibrations of a wire...

Lab 10 Resonant Vibrations of a Wire

Objectives:

< To determine the relation between the frequencies of resonant vibrations of a wire and thephysical parameters of tension, length, and mass per unit length of the wire.

< To determine the relation between the velocity of waves on a wire and the tension andmass per unit length of the wire.

Equipment:

<<<< Pasco Digital Function Generator-Amplifier< Banana leads, 2, 1.5 m long, alligator clips, 2< Large magnet < Table clamps, 2 < Pulley, rods, weight set< Wires with m/L about 0.02, 0.005, 0.002, 0.001, 0.0005, 0.0003 kg/m < Graphical Analysis software Physical Principles: A wave is a disturbance in a medium. Some wave disturbances have the character of a singlepulse. Often waves have a repetitive character. For such waves there is a characteristic shortesttime in which the wave motion repeats. This time is called the period of the motion. Thenumber of oscillations of the wave per second is called the frequency and is equal to 1 / period. There is also a shortest distance in which the wave motion repeats. This distance is called thewave length 8. The relation between frequency f, speed of wave propagation v, and wavelength8, is given by

The velocity of a wave in a string or wire will depend on the tension T and the mass per unitlength (µ = m/L) of the string. When a string under tension is pulled sideways and released, thetension in the string is responsible for accelerating a particular segment back toward itsequilibrium position. The acceleration and wave velocity increase with increasing tension in thestring. Likewise, the wave velocity v is inversely related to the mass per unit length of the string. This is because it is more difficult to accelerate (and impart a large wave velocity) to a massivestring compared to a light string. The exact relationship between the wave velocity v, the tensionT, and the mass per length µ, is given by

If a stretched string is clamped at both ends, traveling waves will reflect from the fixed ends,creating waves traveling in both directions. The incident and reflected waves will combineaccording to the superposition principle. As an example, when the string is vibrated at exactly

Objectives:

Equipment:

Objectives:

Equipment:

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Figure 1 Experimental setup.

Figure 2 Pasco Signal Generator.

the right frequency, a crest moving toward one end and a reflected trough will meet at some pointalong the string. The two waves will cancel at this point, which is called a node. The resultingpattern on the string is one in which the wave appears to stand still, and we have what is called astanding wave on the string. At the nodes, there is no motion in the string. The points whichvibrate with maximum amplitude are called anti-nodes. If the string has length L and is fixed atboth ends, the condition for achieving standing waves is that the length of the string be equal to ahalf-integral number of wavelengths.

(3)L nn

Combining these three results in equations (1), (2) and (3) gives

(4)f vn

Tn = ⋅ =

2λ µ

as the basic model for a resonantly oscillating wire. The lowest or fundamental frequency occurswhen n = 1 and the wave length is 8 = L / 2. The fundamental frequency is given by

. (5)fL

Procedure:

Secure the two table clamps on opposite ends of thetable about 20 cm from the edge. Mount the pulleyand rod in the table clamp away from the wall andclamp the 30 cm rod in the other. Select a wire with amass of about .005 kg/m that is about 2 m long withloops on each end. Support the weight hanger withtotal mass M of about 500 g on the pulley end and place the other loop on the rod. Connect theleads from the 8 S terminals of the Pasco functiongenerator to each end of the wire. Make certain thatthe wire is bare at the ends and any enamel isremoved. Set the generator frequency range switch tothe 1-100 position and the slide switch to the sinewave position. Turn the amplitude all the way up (seefigures 1 and 2).

Part 1: Relationship between resonant frequenciesand the number of nodes

(3)L nn

(4)f vn

Tn = ⋅ =

2λ µ

. (5)fL

Procedure:

(3)L nn

(4)f vn

Tn = ⋅ =

2λ µ

. (5)fL

Procedure:

(3)L nn

(4)f vn

Tn = ⋅ =

2λ µ

. (5)fL

Procedure:

(3)L nn

(4)f vn

Tn = ⋅ =

2λ µ

. (5)fL

Procedure:

From equations (4) and (5) fn = f1 n where f1, the fundamental frequency, will be the slope ofthe graph. Print the graph. Vary the frequency of the signal generator output until the stringvibrates resonantly. Record in Table 1 the frequency f, the number n of anti-nodes (points ofmaximum displacement), and the value f / n. Repeat this process for n increasing from one tofour. Use Graphical Analysis to plot a graph of f vs. n.

Analysis:1. Calculate and record f1 from equation (5).2. Both f / n and the slope should equal the fundamental frequency determined in step 1.

Calculate and record the mean value of f / n.3. Calculate the percentage error between the slope and f1 determined in 1.4. What is the velocity v of the wave in this experiment? Calculate the velocity v two

different ways: First, using equation (2), . Second, using equation (1), v = 8 f1v' T/µwhere 8 = 2 L.

Part 2: Relationship between resonant frequencies and tension

Using the same string from part 1, vary the frequency until resonance occurs for four tensionvalues of between 1 and 15 Newton (hanging masses of M = 0.1, 0.4, 1 and 1.5 kg). Use thesame number of anti-nodes (e.g. three) in each step. Record the values for M, T = M g, fn, ,T

and in Table 2. Use Graphical Analysis to construct a graph of f vs. . v f f L nn= =λ 2 / T

From equation (4),

2 µand the quantity in parenthesis will be the slope of the graph. Print the graph. Also use GraphicalAnalysis to construct a graph of v vs. . From equation 2, T

(7)v T=

µand the quantity in parenthesis will be the slope. Print the graph.

Analysis:

1. Compute the percentage error between the value of the slope calculated from the

expression and the slope of the fn vs. graph.n

L2 µT

2. Compute the percentage error between the value of the slope calculated from theexpression and the slope of the v vs. graph.1/ µ T

For extra credit you may do the following third part.

different ways: First, using equation (2), . Second, using equation (1), v = 8 f1v ' T/µwhere 8 = 2 L.

From equation (4),

(7)v T=

Analysis:

L2 µT

From equation (4),

(7)v T=

Analysis:

L2 µT

From equation (4),

(7)v T=

Analysis:

L2 µT

From equation (4),

(7)v T=

Analysis:

L2 µT

Part 3: Relationship between mass per unit length and the resonant frequency

Holding the tension at about 5N (m = 0.5kg) and using the same number of anti-nodes (e.g.three), select strings with masses per unit length of about 0.02, 0.003, 0.001, 0.0006, 0.0004kg/m. For each of these strings with the same tension and mode number, find the frequency of

resonant vibration. Record the values for µ = m/L, f, , and in a table. Plot a1/ µ v' 2Lfn /n

graph of f vs. in Graphical Analysis. From equation 4, 1/ µ

(8)fn T

µand the quantity in parenthesis will be the slope of the graph. Print the graph. Also use Graphical

Analysis to construct a graph of v vs. . From equation 2, 1/ µ

(9)( )v T=1

Analysis:

expression and the slope of the f vs. graph. n T

L21/ µ

4. Compute the percentage error between the value of the slope calculated from theexpression and the slope of the v vs. graph.T 1/ µ

(8)fn T

(9)( )v T=1

Analysis:

L21/ µ

(8)fn T

(9)( )v T=1

Analysis:

L21/ µ

(8)fn T

(9)( )v T=1

Analysis:

L21/ µ

(8)fn T

(9)( )v T=1

Analysis:

L21/ µ

Recording data:Part 1M = T = M g = L = µ/L =

Table 1 Relationship between resonant frequencies and the number of anti-nodes

n f f / n

slope = % error = 1001

slope f

fPart 2 n = 8 = 2 L / n = µ =

Table 2 Relationship between resonant frequencies and tension

M fn T T v = f 8

slope = f a T an

Ln = = =2 µ

% error = 100 ⋅−

=slope a

slope = v b T b= = =1

% error = 100 ⋅−

=slope b

n f f / n

slope f

fPart 2 n = 8 = 2 L / n = µ =

M fn T T v = f 8

slope = f a T an

Ln = = =2 µ

% error = 100 ⋅−

=slope a

% error = 100 ⋅−

=slope b

n f f / n

slope f

fPart 2 n = 8 = 2 L / n = µ =

M fn T T v = f 8

slope = f a T an

Ln = = =2 µ

% error = 100 ⋅−

=slope a

% error = 100 ⋅−

=slope b

n f f / n

slope f

fPart 2 n = 8 = 2 L / n = µ =

M fn T T v = f 8

slope = f a T an

Ln = = =2 µ

% error = 100 ⋅−

=slope a

% error = 100 ⋅−

=slope b

n f f / n

slope f

fPart 2 n = 8 = 2 L / n = µ =

M fn T T v = f 8

slope = f a T an

Ln = = =2 µ

% error = 100 ⋅−

=slope a

% error = 100 ⋅−

=slope b

Extra Credit Part 3

M = T = M g = n = 8n =

Table 3 Relationship between mass per unit length and the resonant frequency

µ 1µ

fn v = fn 8n

slope = f c cn T

Ln = = =1

% error = 100 ⋅−

=slope c

slope = v d T d= = =1

% error = 100 ⋅−

=slope d

Extra Credit Part 3

M = T = M g = n = 8n =

µ 1µ

fn v = fn 8n

slope = f c cn T

Ln = = =1

% error = 100 ⋅−

=slope c

% error = 100 ⋅−

=slope d

Extra Credit Part 3

M = T = M g = n = 8n =

µ 1µ

fn v = fn 8n

slope = f c cn T

Ln = = =1

% error = 100 ⋅−

=slope c

% error = 100 ⋅−

=slope d

Extra Credit Part 3

M = T = M g = n = 8n =

µ 1µ

fn v = fn 8n

slope = f c cn T

Ln = = =1

% error = 100 ⋅−

=slope c

% error = 100 ⋅−

=slope d

Extra Credit Part 3

M = T = M g = n = 8n =

µ 1µ

fn v = fn 8n

slope = f c cn T

Ln = = =1

% error = 100 ⋅−

=slope c

% error = 100 ⋅−

=slope d

lab 10 resonant vibrations of a wire - andrews …...lab 10 resonant vibrations of a wire...

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