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