A student is investigating the oscillations of a mass attached to two springs connected in series, as shown in Fig. 2.1. springs mass Fig. 2.1 A stopwatch is used to measure the time t for 10 oscillations. The measurement of t is repeated and the average period T is determined. The experiment is repeated for different masses. It is suggested that T and mass M are related by the equation T= (2πM^q) / √k where k is the spring constant of the two springs in series and q is a constant.
- (a)A graph is plotted of lg T on the y-axis against lg M on the x-axis. Determine expressions for the gradient and the y-intercept. gradient = y-intercept =[2]
- (b)Values of M, lg (M/g) and measurements of t are given in Fig. 2.2. M/g t/s t/s T/s Ig (M/g) Ig (T/s) 155 15.2 16.0 2.190 205 18.3 17.5 2.312 250 19.3 20.1 2.398 305 21.0 21.8 2.484 355 23.5 22.7 2.550 410 24.1 24.9 2.613 Fig. 2.2 Calculate and record values of T/s and lg (T/s) in Fig. 2.2. Include the absolute uncertainties in T/s and lg (T/s).[4]
- (c)(i)Plot a graph of lg (T/s) against lg (M/g). Include error bars for lg (T/s).[2]
- (c)(ii)Draw the straight line of best fit and a worst acceptable straight line on your graph. Both lines should be clearly labelled.[2]
- (c)(iii)Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer. gradient =[2]
- (c)(iv)Determine the y-intercept of the line of best fit. Do not include the absolute uncertainty. y-intercept =[1]
- (d)Using your answers to (a), (c)(iii) and (c)(iv), determine the values of k and q. You need not be concerned with units. Do not include the absolute uncertainties. k = q =[2]
- (e)Using your answers to (d), determine the mass M needed to give a period of 1.0s. M = ...g[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [1] gradient = q
- [1] y-intercept = lg(2π/√k) or lg 2π - 1/2 lg k
- [1] Values of T as above. (1.56, 1.79, 1.97, 2.14, 2.31, 2.45)
- [1] Values of lg T as above. (0.193 or 0.1931, 0.253 or 0.2529, 0.294 or 0.2945, 0.330 or 0.3304, 0.364 or 0.3636, 0.389 or 0.3892)
- [1] Uncertainties in T all ±0.04.
- [1] Uncertainties in lg (T/s) consistent with uncertainties in T e.g. from ±0.011 to ±0.007.
- [1] Six points plotted correctly. Must be accurate to the nearest half a small square. Diameter of points must be less than half a small square.
- [1] Error bars in lg T plotted correctly. All error bars must be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
- [1] Line of best fit drawn. Lower end of line should pass between (2.22, 0.22) and (2.25, 0.22) and upper end of line should pass between (2.49, 0.34) and (2.51, 0.34). Do not accept line from first to last plot.
- [1] Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). All error bars must be plotted.
- [1] Gradient determined with clear substitution of data points from the line of best fit into Δy / Δx. Distance between data points must be greater than half the length of the drawn line.
- [1] uncertainty = gradient of line of best fit – gradient of worst acceptable line or uncertainty = 1/2 (steepest worst line gradient – shallowest worst line gradient)
- [1] y-intercept determined by substitution of correct point from the line of best fit into y = mx + c.
- [1] k determined from y-intercept. k = (2π / 10^(y-intercept))^2 = (2π / 10^(c(iv)))^2
- [1] q = answer to (c)(iii) and given to 2 or 3 significant figures.
- [1] M determined from (d) or (c)(iii) and (c)(iv) with correct substitution shown. M = (T√k / 2π)^(1/q) = (T√k / 2π)^(1/(c(iii))) or M = 10^((lg T - y-intercept) / q) = 10^((lg 1 - lg(2π/√k)) / q)
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