Simple harmonic motion
The equation for the acceleration of a body
undergoing simple harmonic motion is usually written as:
acceleration (a) = - ωx
The solution to this equation can be shown to be of
the form
Displacement in s.h.m: x = r sin ωt
since the equation
for acceleration can be obtained by twice differentiating the equation for displacement against
time.
If we differentiate this equation we have alternative equations for both v and
a:
Velocity and acceleration: v = r sin ωt and a = - rω2 sin ωt
We will prove
later that the period of the motion is given by the equation:
Period (T) = 2π/ω
Example problem
An object performs s.h.m. of amplitude 5 cm and period 4 s. If timing is started when the object is at the centre of an oscillation (i.e. x = 0) calculate:
(a) the frequency of the oscillation,
(b) the displacement 0.5 s after the start
(c) the maximum acceleration of the system and
(d) the velocity at a displacement of 3 cm.
(a) Since T= 1/f f =0.25 Hz.
(b) Since ω= 2πf, ω = 1.6 and therefore ωt = 1.6 x 0.5 = 0.8 radians = 45o
Therefore, from x = rsin ωt, x= 5x sin 45o = 3.54 cm
(c) The maximum acceleration occurs when x= r
Therefore maximum acceleration = - ω2r = -1.6 x 1.6 x 5 = -12.8 cms-2
(d) Velocity v is given by the formula
v = +/-ωv(r2 - x2) = +/- 1.6√(52 – 32)
= +6.4cms-1
Note the conversion of radians to degrees in part (b), and the + and - signs in part (d).
A VERSION IN WORD IS AVAILABLE ON THE SCHOOLPHYSICS CD
