A student is investigating stationary waves on a stretched elastic cord. A vibrator attached to the cord is connected to a signal generator. The apparatus is set up as shown in Fig. 2.1. [Figure 2.1] The mass M attached to the cord is adjusted until resonance is obtained. The number n of antinodes on the stationary wave is recorded. The experiment is repeated with different masses to obtain different values of n. It is suggested that M and n are related by the equation f = n / (2L) * sqrt(Mg / μ) where f is the frequency of the vibrator, g is the acceleration of free fall, L is the length of the elastic cord and μ is the mass per unit length of the elastic cord.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

1. [1] gradient = 4d^2f^2/g
2. [1] First mark for uncertainties in first column correct.
3. [1] Second mark for all second column correct.
4. [1] Six points plotted correctly. Must be within half a small square. Diameter of points must be less than half a small square.
5. [1] Error bars in M plotted correctly. All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
6. [1] Line of best fit drawn. Line must not pass through plotted point (0.11, 850) or (0.111, 850).
7. [1] If points are plotted correctly then lower end of line should pass between (0.032, 250) and (0.036, 250) and upper end of line should pass between (0.098, 800) and (0.104, 800).
8. [1] Worst acceptable line drawn (steepest or shallowest possible line). All error bars must be plotted.
9. [1] Gradient determined with a triangle that is at least half the length of the drawn line.
10. [1] uncertainty = gradient of line of best fit – gradient of worst acceptable line or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
11. [1] xdetermined correctly using gradient.
12. [1] x= 9.81 / (4 x 120^2 x 1.54^2) × gradient
13. [1] xdetermined using gradient and given to 2 or 3 significant figures.
14. [1] xdetermined using gradient and correct unit gm⁻¹ and in the range 0.560–0.630 (gm⁻¹).
15. [1] % uncertainty = (2x(0.01 / 1.54) + 2x(5 / 120) + (Δgradient / gradient))x100
16. [1] Correct substitution of numbers must be seen.
17. [1] M determined correctly using x from (d)(i).
18. [1] M = (180^2 x 1.54^2 x (d)(i)) / (9.81×1000) = 7.833 × (d)(i)
19. [1] Correct substitution of numbers must be seen.
20. [1] Absolute uncertainty determined.
21. [1] % uncertainty = (2x(0.01 / 1.54) + 2x(5 / 180) + (d)(ii))x100 = 6.9% + (d)(ii)
22. [1] Correct substitution of numbers must be seen.
23. [1] Maximum/minimum methods: max M = (4x)185^2 x 1.55^2 × max(d)(i) / (4x)9.81×1000 = 8.382 × max(d)(i)
24. [1] min M = (4x)175^2 x 1.53^2 × min(d)(i) / (4x)9.81×1000 = 7.308 × min(d)(i)