Examstyle Questions: Simple Harmonic Motion and Damping

Fig. 2.1 shows an airtrack glider of mass 0.40 Kg held by two strecthed strings. When the glider is pulled 0.050 m to the left and released, it oscillates freely without friction.
Fig. 2.2 shows the variation of the elastic strain energy stored in the springs with the displacement x from the equilibrium posotion. Note that the strain energy is 70 mJ when the glider is not oscillating.
a) Write down
i) the total energy stored in the system when oscillating ..................................... mJ
(1 Mark)
ii) the maximum kinetic energy of the glider .................................... mJ
(2 Marks)
b) i) Show that the maximum speed of the glider is 0.50 ms1.
(2 Marks)
ii) Use Fig. 2.2 or otherwaise to find the amplitude of oscillation required to halve the maximum speed of the glider. Show your reasoning.
amplitude = ......................... m
(2 Marks)
c) The equation of motion of the glider relating its acceleration a in m s ^{2} to its displacement x in m is
a = 110 x
i) Use the equation to show that the period of ocsillation is 0.60 s.
(2 Marks)
ii) Use the data from (b)(i) and (c)(i) to sketh on Fig. 2.3 the velocitytime graph for the glider. It is released at x = 0.050 m at t = 0.
(3 Marks)
(Marks available: 12)

The variation in depth of water in a harbour can be modelled as a simple harmonic oscillation.
Fig. 7.1 shows a graph produced fom the model. It shows the variation of depth of water in a harbour with time over the period of one day (24 hours).
a) Use the grpah to find
i) the maximum depth of the water in the harbour
maximum depth = .................. m
(1 Mark)
ii) the amplitude A of the tidal motion
amplitude of motion = ....................... m
(1 Mark)
iii) the rate of change of deptg in metres per hour (m hr1) at t = 6 hours where t = time after midnight in hours. Show your working.
rate of change in depth = ................. m hr^{1}
(3 Marks)
b) Use data from the graph to calculate the frequency f of the tidal motion in units of tides per hour
frequency = ..................... tides per hour
(3 Marks)
c) The equation for the depth of water d, in metres, in the harbour is
d = 10 + A sin (2π f t).
Use your answer to a)(ii) and this equation to show that the lowest depth of water is 5 m.
(2 Marks)
(Marks available: 10)
Answer outline and marking scheme for question:

a) i) 120 (mJ)
(1 Mark)
ii) 120 70 ; = 50(mJ) give 2 marks for correct answer without working
(2 Marks)
b) i) k.e. = ½ mv^{2} = 50 x 10^{3} = 0.2 V^{2} (ecf from a(ii))
v^{2} = 0.25 ; v = 0.5 (m s^{1})
(2 Marks)
ii) reasoning, e.g. max energy = ½ mv^{2} = ½ kA^{2} so A proportional to V_{m}/AW;
or max ke  12.5 mJ so total energy = 8.25 mJ, read x from graph;
giving A = 0.025 (m)
(2 Marks)
c) i) a = 4π^{2}f^{2}x;
f^{2} = 110/ 4π^{2} = 2.786 / f = 10.5 / 2π = 1.67 so f = 1/T = 0.6 (s)
(2 Marks)
ii) sinusoidal wave with correct period; correct amplitude correct phase accept A or A at 0.15 s
(3 Marks)
(Marks available: 12)

a) i) Max depth = 15 m
(1 Mark)
ii) Amplitude = 5 m
(1 Mark)
iii) Gradient at t = 6 hours (1 Mark) correct reading from graph (1 Mark) answer worked to 3.0 m hr^{1} (1 Mark) (answers in range 2.5 m hr^{1} to 3.5 m hr^{1})
(3 Marks)
b) time period from graph = 12.5 hrs (1 Mark) f = 1/T = 1/12.5 (1 Mark) = 0.050 hr^{1 }(1 Mark)
(3 Marks)
c) d = 10 + 5 sin (2πsin 0.080 * 9.5) (1 Mark) = 5.0 m (1 Mark) (allow ecf from a(ii) and b) or sin varies between +1 and 1 (1 Mark) so lowest value is 10A (this allows incorrect value for A to ecf) (1 Mark)
(2 Marks)
(Marks available: 10)