Learning objective one, due February 1st 2015.

1. Periodic motion in a
simple pendulum

2. The Problem
A object of mass 5kg is hanging from an inextensible string of length
2m attached to a frictionless rigid support. It is displaced by an
angular distance D forming an angle x of 45 degrees. Find
(a) (i) The displacement D in meters
(ii) The acceleration at
(b) Assuming that the initial angle of displacement were to be reduced
so as to be very small, find angular frequency W. Would reducing the
angel cause the acceleration to increase, decrease or stay the same?
Justify your reason with equations.

3. Visualize the problem
Photo taken from “Physics
for Scientists and
Engineers An Interactive
Approach revised custom
volume 1 PHYS 101” page
172
HAWKES
IQBAL
MANSOUR
MILNER-BOLTON
WILLIAMS
D

4. Understanding the problem
 Let us consider the case if the angel of displacement were = 0 degrees. From here
we can see that T=mg because there is no movement along the axis of the string.
 This also means that as the string is displaced T=mgCosx
 As the angle x increases the magnitude of the weight acting in the opposite
direction to the tension (mgCosx) decreases towards zero, this is intuitive, if a
string was held completely horizontally outwards, we wouldn’t expect there to be
any tension.
 Applying the same logic, looking at the diagram if we were to take a tangent of the
arc formed by displacement D, the magnitude of the weight component would
equal mgSinx. But, unlike the tension force there is no other component equal and
opposite to it, so there is a net force.
 Fnet= (negative) mgSinx=mat
 The sign is negative because from the diagram mgSinx is acting to the left, which
we have taken to be the negative direction
 The mass cancels out and we are left with at = (negative) gSinx

5. Solving the problem
(a) (i) If we consider the displacement D as the arc of a circle
relative to the equilibrium position. D=L*x (in radians). So the
displacement = 2*pi/4
Therefore D = 1.57m
(ii) We have found our expression for acceleration in
“Understanding the problem” now we just need to plug in the
values
at= (negative) gSinx = (negative) 9.8*Sin(pi/4)
= 6.93m/s2
I of course put the negative in brackets to show that the sign of the
acceleration is dependent upon the direction of displacement. If the pendulum
had been displaced in the opposite direction as shown in the diagram the
sign would have been positive.

6. (b) If you take your calculator and plug in values of x (in radians) you
will see that the smaller x is the closer Sinx comes to approximating it. The
problem specifies a very small value of x so we can assume Sinx = x.
at= (negative) gx. This means that the acceleration of the
mass is proportional to the displacement and opposite in sign, so
this is simple harmonic motion
But we know from problem (a)(i) that D=L*x, so x=D/L
at= (negative) g(D/L) = (negative) (g/L)D.
If you are familiar with simple harmonic motion this looks
familiar, a constant that determines acceleration over a fixed
value. In Simple harmonic motion W= root(k/m).
Applying that same logic here W= root(g/L)= root(9.8/2) =
2.21rad s-1
From intuition we can tell that if x is reduced at will decrease as
well, using equations, at= (negative) g(D/L). Reducing x will result in a
reduction in D, therefore at will also decrease