In: Physics
Data for Experiment 108- Frequency of Vibration
Frequency and Tension
Trial 1 Trial 2
Vibrating Length (L) 50 cm 65 cm
Number of Segments (n) 2 2
Attached Mass (m) 70g 90g
Linear Mass Density 0.0015 g/cm 0.0015 g/cm
Actual Frequency 120 Hz 120 Hz
Find the Following. Use the correct number of Significant Figures.
1. Tension ( T=mg) in Trial 1 and Trial 2
2. Experimental Frequency in both trials.
3. Percentage error in both trials.
Here we will use the fallowing relation between Frequency ( )Tension ( ), linear density of string ( ), Length of the string ( ), and number of segments ( ) ( that is the number/order of harmonic ) ,
______________________________________________ relation 1
let = mass attached to the string and
Trial 1)
after adjusting for significant figures
Trial 2)
after adjusting for significant figures
Table 1
Trial 1 | Trial 2 | |
Vibrating length ( ) ( cm ) |
50 | 65 |
Number of segments ( ) |
2 | 2 |
attached mass ( ) ( g ) |
70 | 90 |
Linear mass density ( ) ( g/cm ) |
0.0015 | 0.0015 |
Tension in string ( ) ( ) |
68000 | 88000 |
Actual frequency ( ) ( Hz ) |
120 | 120 |
Calculated frequency ( Hz ) |
135.25 | 117.97 |
Calculated frequency in significant digits ( ) ( Hz ) |
||
Percentage error ( ) |
If you have measured string lengths in mm accuracy that is
Trial 1 L = 50.0 cm and Trial 2 L = 65.0 cm
Then last two rows of Table 1 will be as below
Trial 1 | Trial 2 | |
Calculated frequency in significant digits ( ) ( Hz ) |
135 | 118 |
Percentage error ( ) |
12.5 % | 1.67% |
And If you have measured string lengths in only in cm accuracy that is
Trial 1 L = 50 cm and Trial 2 L = 65 cm
Then last two rows of Table 1 will be as below
Trial 1 | Trial 2 | |
Calculated frequency in significant digits ( ) ( Hz ) |
140 | 120 |
Percentage error ( ) |
13.33 % | 0% |