This was a popular question among the responding candidates. Majority had good results from the timing procedure. However, h was not recorded to 1d.p in cm and h ½ was not evaluated to 3 s.f. Some candidates could not recall that h ½ = vh . Many candidates determined the slope, intercept and s-2 correctly. Precaution were also well stated. Part (b) (i) was poorly attempted candidates were able to handle while (ii) was satisfactorily attempted. Candidates are expected to: Measure and record five values of h to 1 d.p in cm Read and record five values of t to 1 d.p in seconds Evaluate five values of T correctly to at least 2 d.p Evaluate five values of h½ correctly to at least 2 d.p Make composite table showing h,t, T and h½ Plot a graph using reasonable scales Draw line of best fit Evaluate the slope s and s-2 of the graph Determine the intercept on the vertical axis State any two of the following precautions in acceptable language Ensured retort stand is firmly clamped Avoided parallax error on stopwatch/clock/metre rule Noted and corrected/avoided zero error on stopwatch/clock/metre rule Avoided draught Repeated readings shown on table Answer part (b) as stated below: b(i) Amplitude is the maximum displacement of an oscillating object from an equilibrium position. (ii) F = m(1)2r v = (1)2r = 2nr/T = m( 2p/T )2r OR = 2 x 22/7 x0.7/0.5 = 8.8 ms-1 = 0.5(2 x 22/7 x 1/0.5 )2x 0.7 F = mv²/r = 55.3N = (0.5 x (8.8)²)/0.7 = 55.3 N