2005 WAEC Physics Practical You have been provided with a retort stand, clamp and boss, a set of masses,...

WAEC 2005

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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.

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Explanation

Table of values/observation
Number of oscillations (n) = 20

S/N	M(g)	t\(_{1}\)(s)	t\(_{2}\)(s)	t = \(\frac{t_{1}+t_{2}}{2(s)}\)	T = \(\frac{t}{n}\)	T\(^{2}\)(sec\(^{2}\))	T\(^{2}\)(sec\(^{2}\))x10\(^{-3}\)
1	50.0	5.80	5.80	5.80	0.290	0.084	84
2	70.0	6.70	6.90	6.80	0.340	0.116	116
3	90.0	7.70	7.90	7.80	0.395	0.156	156
4	110.0	8.50	8.70	8.60	0.430	0.185	185
5	130.0	9.20	9.20	9.20	0.460	0.212	212

(7)The graph of T\(^{2}\)(sec\(^{2}\)) Vs m(g) axis

Scale: T\(^{2}\)(sec\(^{2}\)) axis: 2cm = 20 units

m(g) axis: 2cm = 20 units

No description available.

(8) Slope/gradient = \(\frac{\bigtriangleup {T}^{2}(sec^{2})}{\bigtriangleup {m(s)}}\)

\(\frac{234-10}{140-6} = \frac{224}{134}\)

= 1.672 sec\(^{2}\)/g

l on the vertical axis is (0)

(9) K = \(\frac{4\pi ^{2}}{s} = \frac{4 \times ({\frac{22}{7})^{2}}}{1.672}\)

= \(\frac{4 \times (22)^{2}}{7^{2} \times 1.672} = \frac{4 \times 484}{49 \times 1.672}\)

= \(\frac{1936}{81.928}\) = 23.61 g/sec\(^{2}\)

Precautions are:

I ensured that I took care of zero error when taking the period of oscillation.
I ensured that there was no conical oscillation when the mass hanger was loaded with different slotted masses.
I ensured that I took repeated readings in order to avoid random errors.

(b)i. Young modulus is defined as the ratio of tensile or Compressive stress to tensile or compressive strain. It is denoted by E and measured in N/m\(^{2}\). Thus,

Young modulus = \(\frac{\text {tensile/compressive stress}}{\text{tensile/compressive strain}}\)

E = \(\frac{F/A}{e/L} =\frac{FL}{Ae}\)

Force constant is defined as the amount of force that causes a unit extension of an elastic material or the ratio of F/e of an elastic material. It is denoted by K and measured in N/m or N/m\(^{-1}\). Thus,

F = Ke, K = \(\frac{f}{e}\)

ii. F = 500N, K = 1.0 x 10\(^{4}\) Nm\(^{-1}\), F = ke

500 = 1 x 10\(^{4}\) x e

e = \(\frac{500}{1 \times 10^{4}} = \frac{5 \times 10^{2}}{1 \times 10^{4}}\) = 5 x 10\(^{2-4}\)

= 5 x 10\(^{-2}\) (0.05)m

W = ½Fe = ½ x 500 x 0.05 = 12.5J

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