You have been provided with a retort stand, clamp and boss, a set of masses, a spiral spring, stopwatch, split cork, and other necessary apparatus. Using th...
Question 1 Report
You have been provided with a retort stand, clamp and boss, a set of masses, a spiral spring, stopwatch, split cork, and other necessary apparatus. Using the diagram above as a guide, carry out the following instructions;
- Suspend the spiral spring vertically as shown in the diagram.
- Suspend a mass hanger on the free end of the spiral spring and add a mass, m 50.0g to the hanger.
- Pull the hanger gently downwards and release to set it into vertical 0scillations.
- Determine the time, t, for 20 complete oscillations.
- Evaluate the period, t, of the oscillation. Also, evaluate T\(^{2}\).
- Repeat the procedure for four other values of m= 70, 90, 110, and 130g. In each case, determine t and evaluate T and T\(^{2}\). Tabulate your readings.
- Plot a graph of T\(^{2}\) on the vertical axis against m on the horizontal axis.
- Determine the slope, s, of the graph and the intercept, l, on the vertical axis.
- Evaluate k = 4\(\frac{\pi ^{2}}{s}\), Take t = \(\frac{22}{7}\)
- State two precautions taken to ensure accurate results.
(b)i. Define Young modulus and force constant.
ii. A force of magnitude 500N is applied to the free end of a spiral spring of force constant 1.0 x 10\(^{4}\) Nm\(^{-1}\). Calculate the energy stored in the stretched spring.
Answer Details
Observation and calculations
For each load, the time for 20 complete oscillations was measured twice. The mean time is \(t\), hence
\[T=\frac{t}{20},\qquad T^2=\left(\frac{t}{20}\right)^2.\]
| Mass, \(m\) (g) | \(t_1\) (s) | \(t_2\) (s) | Mean \(t\) for 20 oscillations (s) | Period, \(T=t/20\) (s) | \(T^2\) (s2) | \(T^2\times10^3\) (s2) |
|---|---|---|---|---|---|---|
| 50.0 | 5.80 | 5.80 | 5.80 | 0.290 | 0.084 | 84 |
| 70.0 | 6.70 | 6.90 | 6.80 | 0.340 | 0.116 | 116 |
| 90.0 | 7.80 | 8.00 | 7.90 | 0.395 | 0.156 | 156 |
| 110.0 | 8.50 | 8.70 | 8.60 | 0.430 | 0.185 | 185 |
| 130.0 | 9.20 | 9.20 | 9.20 | 0.460 | 0.212 | 212 |
The graph is plotted with \(T^2\times10^3\) on the vertical axis and mass on the horizontal axis.
Gradient, intercept and force constant
Using two widely separated points on the straight line, \((m,T^2\times10^3)=(6,10)\) and \((140,234)\),
\[s=\frac{(234-10)\times10^{-3}}{140-6}=1.672\times10^{-3}\ \text{s}^2\text{g}^{-1}=1.672\ \text{s}^2\text{kg}^{-1}.\]
The vertical intercept, \(l\), is approximately \(0\ \text{s}^2\).
Since \(T^2=\dfrac{4\pi^2}{k}m\),
\[k=\frac{4\pi^2}{s}=\frac{4\left(\frac{22}{7}\right)^2}{1.672}=23.6\ \text{N m}^{-1}.\]
Precautions
- The load was made to oscillate vertically only; conical or sideways oscillations were avoided.
- Small oscillations were used, and the time for 20 oscillations was measured and repeated before taking the mean.
(b)(i) Definitions
Young modulus is the ratio of tensile stress to tensile strain, provided the elastic limit is not exceeded:
\[E=\frac{F/A}{e/L}=\frac{FL}{Ae}.\]
Force constant is the force required to produce unit extension of a spring:
\[k=\frac{F}{e}.\]
(b)(ii) Energy stored in the spring
\[e=\frac{F}{k}=\frac{500}{1.0\times10^4}=0.050\ \text{m}.\]
\[W=\frac{1}{2}Fe=\frac{1}{2}\times500\times0.050=12.5\ \text{J}.\]
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