You are provided with two retort stands, two-metre rules, pieces of thread and other necessary apparatus. i. Set up the apparatus as illustrated above ensur...
You are provided with two retort stands, two-metre rules, pieces of thread and other necessary apparatus.
i. Set up the apparatus as illustrated above ensuring the strings are permanently 10cm from either end of the rule.
ii. Measure and record the length L = 80 cm of the two strings.
iii. Hold both ends of the rule and displace the rule slightly, then release so that it oscillates about a vertical axis through its centre.
iv. Determine and record the time t for 10 complete oscillations.
v. Determine the period T of oscillations.
vi. Evaluate log T and L.
vii. Repeat the procedure for four other values of L= 70 cm, 60 cm, 50 cm, and 40 cm
viii. Tabulate your readings.
ix. Plot a graph with log T on the vertical axis and log L on the horizontal axis.
x. Determine the slope, s, and the intercept, c on the vertical axis.
xi. State two precautions taken to ensure accurate results.
(b)i. Define simple harmonic motion.
ii. Determine the value of L corresponding to t= 12 s from the graph in 1.
Bifilar pendulum: torsional oscillation of a suspended metre rule
The two threads of equal length \(L\) are fixed to the rigid horizontal support, each 10 cm from the ends of the metre rule, so that the rule hangs horizontally and can oscillate about the vertical axis through its centre.
Bifilar pendulum: metre rule suspended by two threads of length L, each 10 cm from the ends, oscillating about the vertical axis through its centre.
For each length the period is obtained from the timing of ten complete oscillations:
\[ T = \frac{t}{10} \]
and \(\log T\) and \(\log L\) are then evaluated for each reading.
(viii) Table of readings
S/N
L /cm
t /s (10 osc.)
T = t/10 /s
log T
log L
1
80.0
17.9
1.79
0.253
1.903
2
70.0
16.7
1.67
0.223
1.845
3
60.0
15.5
1.55
0.190
1.778
4
50.0
14.1
1.41
0.149
1.699
5
40.0
12.6
1.26
0.100
1.602
(ix) Graph of log T against log L
Straight-line graph of log T (vertical) against log L (horizontal); slope s = 0.51, intercept c = -0.71.
The points lie on a straight line, confirming that \( \log T = s\,\log L + c \).
(x) Slope and intercept
Taking two widely separated points on the line of best fit, \((1.602,\;0.100)\) and \((1.903,\;0.253)\):
Extending the line back to \(\log L = 0\) (or using \( c = \log T - s\log L = 0.253 - 0.51\times1.903 \)) gives the vertical intercept:
\[ c = -0.71 \]
Hence \( \log T = 0.51\,\log L - 0.71 \), which corresponds to \( T \propto L^{1/2} \), the expected law for the bifilar pendulum.
(xi) Two precautions
I avoided parallax error when reading the metre rule and when starting and stopping the stopwatch, by viewing each scale directly from the front.
I ensured that the support was rigid and that the rule oscillated smoothly about a vertical axis through its centre with only a small angular displacement, counting the oscillations from a fixed reference mark.
(b)(i) Simple harmonic motion
Simple harmonic motion is the motion of a body whose acceleration is directly proportional to its displacement from a fixed point and is always directed towards that fixed point:
\[ a = -\omega^{2}x \]
(b)(ii) Value of L corresponding to t = 12 s
For \( t = 12\,\text{s} \):
\[ T = \frac{t}{10} = \frac{12}{10} = 1.2\,\text{s}, \qquad \log T = \log 1.2 = 0.079 \]
Reading from \( \log T = 0.079 \) on the vertical axis across to the line of best fit and down to the horizontal axis (or solving \( 0.079 = 0.51\log L - 0.71 \)):
\[ \log L = \frac{0.079 + 0.71}{0.51} = \frac{0.789}{0.51} = 1.56 \]\[ L = 10^{1.56} = 36\,\text{cm} \]
Therefore the length of the threads corresponding to \( t = 12\,\text{s} \) is \( L \approx 36\,\text{cm} \).
Bifilar pendulum: torsional oscillation of a suspended metre rule
The two threads of equal length \(L\) are fixed to the rigid horizontal support, each 10 cm from the ends of the metre rule, so that the rule hangs horizontally and can oscillate about the vertical axis through its centre.
Bifilar pendulum: metre rule suspended by two threads of length L, each 10 cm from the ends, oscillating about the vertical axis through its centre.
For each length the period is obtained from the timing of ten complete oscillations:
\[ T = \frac{t}{10} \]
and \(\log T\) and \(\log L\) are then evaluated for each reading.
(viii) Table of readings
S/N
L /cm
t /s (10 osc.)
T = t/10 /s
log T
log L
1
80.0
17.9
1.79
0.253
1.903
2
70.0
16.7
1.67
0.223
1.845
3
60.0
15.5
1.55
0.190
1.778
4
50.0
14.1
1.41
0.149
1.699
5
40.0
12.6
1.26
0.100
1.602
(ix) Graph of log T against log L
Straight-line graph of log T (vertical) against log L (horizontal); slope s = 0.51, intercept c = -0.71.
The points lie on a straight line, confirming that \( \log T = s\,\log L + c \).
(x) Slope and intercept
Taking two widely separated points on the line of best fit, \((1.602,\;0.100)\) and \((1.903,\;0.253)\):
Extending the line back to \(\log L = 0\) (or using \( c = \log T - s\log L = 0.253 - 0.51\times1.903 \)) gives the vertical intercept:
\[ c = -0.71 \]
Hence \( \log T = 0.51\,\log L - 0.71 \), which corresponds to \( T \propto L^{1/2} \), the expected law for the bifilar pendulum.
(xi) Two precautions
I avoided parallax error when reading the metre rule and when starting and stopping the stopwatch, by viewing each scale directly from the front.
I ensured that the support was rigid and that the rule oscillated smoothly about a vertical axis through its centre with only a small angular displacement, counting the oscillations from a fixed reference mark.
(b)(i) Simple harmonic motion
Simple harmonic motion is the motion of a body whose acceleration is directly proportional to its displacement from a fixed point and is always directed towards that fixed point:
\[ a = -\omega^{2}x \]
(b)(ii) Value of L corresponding to t = 12 s
For \( t = 12\,\text{s} \):
\[ T = \frac{t}{10} = \frac{12}{10} = 1.2\,\text{s}, \qquad \log T = \log 1.2 = 0.079 \]
Reading from \( \log T = 0.079 \) on the vertical axis across to the line of best fit and down to the horizontal axis (or solving \( 0.079 = 0.51\log L - 0.71 \)):
\[ \log L = \frac{0.079 + 0.71}{0.51} = \frac{0.789}{0.51} = 1.56 \]\[ L = 10^{1.56} = 36\,\text{cm} \]
Therefore the length of the threads corresponding to \( t = 12\,\text{s} \) is \( L \approx 36\,\text{cm} \).
