Question 1
(a) State what is meant by simple harmonic motion. [2]
(b) A small ball rests at point P on a curved track of radius r, as
shown in Fig. 4.1.
Fig. 4.1
The ball is moved a small distance
to one side and is then released. The horizontal displacement x of the ball is
related to its acceleration a towards P by the expression
a = − gx / r
where g is the acceleration of free
fall.
(i) Show that the ball undergoes
simple harmonic motion. [2]
(ii) The radius r of curvature of
the track is 28 cm.
Determine the time interval Ï„
between the ball passing point P and then returning to point P. [3]
(c) The variation with time t of the displacement x of the ball in (b)
is shown in Fig. 4.2.
Fig. 4.2
Some moisture now forms on the
track, causing the ball to come to rest after approximately 15 oscillations.
On the axes of Fig. 4.2, sketch the
variation with time t of the displacement x of the ball for the first two
periods after the moisture has formed. Assume the moisture forms at time t = 0.
[3]
Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q4
Solution 1:
(a) For a simple harmonic motion, the acceleration / force is proportional
to the displacement (from a fixed point) and the
EITHER acceleration and displacement
are in opposite directions OR
acceleration is always directed towards a fixed point
(b)
(i) Acceleration due to gravity, g
and radius r are constant, so the acceleration a is proportional to the
displacement x.
The negative sign shows that a and x
are in opposite directions.
(ii)
{For simple harmonic
motion, acceleration a = – ω2x
a = – ω2x = –
gx / r giving}
ω2 = g / r and ω = 2π / T
ω2 = 9.8 / 0.28 = 35
T = {2π / ω2
=} 2Ï€ / √35 = 1.06 s
{The ‘T’ calculated above
is the period. The question asks for only the time interval Ï„ between the ball
passing point P and then returning to point P. This is half the period T.}
time interval Ï„ = {T/2 =} 0.53 s
(c)
sketch: time period constant (or
increases very slightly)
drawn line always ‘inside’ given
loops
successive
decrease in peak height
