standing waves report

Date:20-November-2006 Physics 210
STANDING WAVES ON A STRETCHED STRING

A) Resonant frequencies versus n:

Distance between bridges = 0.6 m. Tension = Mg = 9.8 N.
Mode
Signal Generator Frequency (Hz)
Wire frequency (Hz)

Node to node distance (m)
Wavelength (m)
First

39
78
54.9
109.8
Second

75.9
151.8
27.8
55.6
Third

117
234
18.3
36.6
Fourth

160.4
320.8
14
28
Fifth
198.95

397.9
10.1
20.2
Note:
The Wire frequency is double the Signal Generator Frequency
Node to node distance where λ is the wavelength/
Plot the frequency versus n? Deduce from the slope of your straight line the linear density, , of the string (show procedure and calculations). Comment.

Y – axis
X – axis
Frequency (Hz)
N
78
1
151.8
2
234
3
320.8
4
397.9
5

Using the calculator we perform linear regression to obtain the slope B as 80.08 while A = -4.54
Which is nothing but
To obtain the error on the slope: αB

Where N = 5

The error αμ is:

Hence the linear density is 1.1x103 g/m ± 3x10-5 g/m

The actual linear density is 1120 kg/m which means the error is approximately 1.8%

B) Fundamental frequency versus tension

Distance between bridges =0.6 m.

Tension (N)
Fundamental Frequency (Hz)
Wave Velocity (m/s)
9.8
√T1=3.13
78
93.6
19.6
√T2=4.27
109
130.8
29.4
√T3=5.42
135
162
39.2
√T4=6.26
156
187.2
49
√T5=7
174.4
209.3

Plot the frequency versus (T)1/2. Deduce from the slope of your straight line the linear density, , of the string (show procedure and calculations). How does your values compare to that measured in Part A? Comment.

f=B√T + A
The slope B=24.65 & A=1.88 (After performing the calculations on calculator)

αB2 = (N/N-2) (∑ ei2)/Δ (Error on slope)

f1calc = B√T1 + A = (24.65) (3.13) + 1.88= 75.27 Hz f2 calc = (24.65)