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Magazine Chapter 1: Q14-91P (page 1)
Experimenting with pendulums, you attach a light string to the ceiling and attach a small metal sphere to the lower end of the string. When you displace the sphere 2.00 m to the left, it nearly touches a vertical wall; with the string taut, you release the sphere from rest. The sphere swings back and forth as a simple pendulum, and you measure its period T. You repeat this act for strings of various lengthsL, each time starting the motion with the sphere displaced 2.00 m to the left of the vertical position of the string. In each case the sphere鈥檚 radius is very small compared with L. Your results are given in the table:
L (m) | 12.00 | 10.00 | 8.00 | 6.00 | 5.00 | 4.00 | 3.00 | 2.50 | 2.30 |
T (s) | 6.96 | 6.36 | 5.70 | 4.95 | 4.54 | 4.08 | 3.60 | 3.35 | 3.27 |
(a) For the five largest values of L, graph \({T^2}\) versus L. Explain why the data points fall close to a straight line. Does the slope of this line have the value you expected? (b) Add the remaining data to your graph. Explain why the data start to deviate from the straight-line fit as L decreases. To see this effect more clearly, plot \({T \mathord{\left/ {\vphantom {T {{T_0}}}} \right.} {{T_0}}}\) versus L, where \({T_0} = 2\pi \sqrt {{L \mathord{\left/ {\vphantom {L g}} \right.} g}} \) and \(g = 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\). (c) Use your graph of \({T \mathord{\left/ {\vphantom {T {{T_0}}}} \right.} {{T_0}}}\) versus L to estimate the angular amplitude of the pendulum (in degrees) for which the equation \(T = 2\pi \sqrt {{L \mathord{\left/ {\vphantom {L g}} \right. } g}} \) is in error by 5%.
Short Answer
(a) (i) The graph is drawn below. (ii) \(3.913\;{{\rm{s}}^{\rm{2}}}{\rm{/m}}\)
Step by step solution
Given Data
L (m) | 12.00 | 10.00 | 8.00 | 6.00 | 5.00 | 4.00 | 3.00 | 2.50 | 2.30 |
T (s) | 6.96 | 6.36 | 5.70 | 4.95 | 4.54 | 4.08 | 3.60 | 3.35 | 3.27 |
Concept
When you draw a graph between L versus T, the graph will be a curved one.
When you draw a graph between \({T^2}\) versus L, it will be a straight line.
Step 3(a) (i): Plot a graph \({T^2}\) versus L
The graph is plotted between\({T^2}\;{\rm{vs}}\;L\).
L (m) | 12.00 | 10.00 | 8.00 | 6.00 | 5.00 | 4.00 | 3.00 | 2.50 | 2.30 |
\({T^2}\;({{\rm{s}}^2})\) | 48.44 | 40.45 | 32.49 | 24.50 | 20.61 | 16.64 | 12.96 | 12.25 | 10.69 |
Hence the graph is drawn.
Step 3(a) (ii): Find the slope value
The graph shows the equation is,
\({T^2} = 3.877L + 1.586\)
This is the equation of straight line so that the data points fall close to straight line.
In above equation compared to general equation of straight line \(y = mx + c\) is,
\(m = 3.877\;{{\rm{s}}^{\rm{2}}}{\rm{/m}}\)
Formula to calculate the time period of a single pendulum is,
\(\begin{aligned}T &= 2\pi \sqrt {\frac{L}{g}} \\{T^2} &= \frac{{4{\pi ^2}}}{g}L\\{\rm{Where,}}\;L = {\rm{length}}\;{\rm{of}}\;{\rm{pedulum}}\\\;\;\;\;\;\;\;\;\;\;T = {\rm{time}}\;{\rm{period}}\;{\rm{of pendulum}}\end{aligned}\)
So a graph between \({T^2}\;{\rm{and}}\,\;L\) of pendulum gives a straight line with a slope of \(\frac{{4{\pi ^2}}}{g}\)
Substitute \(g = 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\) to find \(\frac{{4{\pi ^2}}}{g}\)
\(\begin{aligned}\frac{{4{\pi ^2}}}{g} &= \frac{{4{\pi ^2}}}{{9.8}}\\ &= 4.02\;{{\rm{s}}^{\rm{2}}}{\rm{/m}}\end{aligned}\)
The value of slope is close to the \(3.913\;{{\rm{s}}^{\rm{2}}}{\rm{/m}}\) so which is fairly close to the expected value.
Hence the slope of the equation is \(3.877\;{{\rm{s}}^{\rm{2}}}{\rm{/m}}\)
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