MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
W15D1-3 Table Problem Coupled Oscillators
A simple pendulum consists of a point mass m attached to the end of a rod of negligible
mass and length l . A spring of negligible mass and force constant k is connected at one
end to the point mass and attached to a wall at the other end. The spring is relaxed when
  0 . The pendulum is displaced a small angle  0 from the vertical and released from
rest. The system oscillates. Let g denote the magnitude of gravitational acceleration.
Express all your answers in terms of g , l , m , k , and  0 as needed. You may use the
small angle approximations for sin and cos . You may also approximate the direction
of the spring force as horizontal throughout the motion.
a) Find the differential equation that describes the relation between the angular
displacement  and the angular acceleration d 2 / dt 2 .
b) Based on your result from part a), write down an expression for the angular
frequency 0 of the oscillation? Hint: Check that your answer makes sense when
either k  0 or g  0 . In these cases 0 should agree with the results you know.
c) What is the solution of the equation of motion you found in part a) for the angular
displacement  (t) ? Assume    0 at t  0 and d / dt  0 at t  0 .
Solution: We begin by choosing a coordinate system and draw the system when it is
stretched and compare it to when it is unstretched as shown in the figure below.
If we only stretch the spring so that the angle  is small then we can approximate the
system by
The amount the spring has stretched is then approximately given by x  l sin  . So the xcomponent of the spring force is kx  kl sin 
The horizontal and vertical component equations of Newton’s Second Law are
2
d x
iˆ :  T sin   kl sin   m 2 .
dt
ˆj : T cos   mg  0
Using the small angle approximations that x / l  sin    , hence d 2 x / dt 2  ld 2 / dt 2
and cos  1 , these equations become
d
iˆ :  T  kl  ml 2
dt
ˆj : T  mg  0 .
2
Thus setting T  mg , the horizontal force equation becomes
d 2
 (mg  kl )  ml 2 ,
dt
which we can rewrite in the form of the simple harmonic oscillator equation
d 2
g k
 (  ) .
2
dt
l m
The angular frequency of oscillation is approximately
0 
g k
 .
l m
Note that when g  0 , 0  k / m which is the angular frequency of a spring-object
system and when k  0 , 0  g / l which is the angular frequency of a simple
pendulum. With the initial condition that the object is released from rest at a small angle
0 , the angle the string makes with the vertical is given
 g k 

t  .
 l m 
 (t )   0 cos 0 t    0 cos 
(0.1)
The component of the angular velocity of the bob is
 g k 
d
g
(t )    0 sin 

t  .
dt
l
l
m


(0.2)
Energy Method: We could use energy considerations as we did for the simple
pendulum. For small angles  with respect to the vertical, we can use the small angle
approximations that cos   1   2 / 2 and sin    , the gravitational potential energy
(relative to a choice of zero potential energy at the bottom of the swing where   0 ) is
given by
U g  mgl 1  cos   
1
mgl 2 .
2
(0.3)
The spring potential energy is approximately
Us 
1 2 1 2 2
1
kx  kl sin   kl 2 2 .
2
2
2
So the total potential energy is approximately
U
1
1
1
g k 
mgl 2  kl 2 2  ml 2     2 .
2
2
2
 l m
The component of the velocity of the object is given by v  l
1
1  d 
K  m v 2  m  l

2
2  dt 
d
so the kinetic energy is
dt
2
The energy of the system is then
1  d  1 2  g k  2
ml
  ml     .
2  dt  2
 l m
2
E  K U 
Since there is no non-conservative work, the energy is constant hence
2
dE 1
1
 g k  d
2 d d 
0
 m 2l
  2ml 2    
2
dt 2
dt dt
2
 l m  dt
d  d 2  g k  
 ml
    

dt  dt 2  l m  
.
2
There are two solutions to this equation, the first one d / dt  0 is the equilibrium
solution, the angular speed is zero means the suspended object is not moving. The second
solution is the one we are interested in
d 2  g k 
     0 .
dt 2  l m 
which is the same differential equation we found using the force method.
Note: Let x0 be the length of the unstretched spring. Then the length x of the stretched
spring is given by
x 
l (1  cos )   l sin   x0 
2
2
Expanding we have
x 
l
2
 l 2 cos 2   2l 2 cos     l 2 sin 2   2l sin  x0  x0 2 
2
Use the trigonometric identity that sin 2   cos 2   1 in the above expression which
simplifies to
x 
 2l
2
(1  cos  )  2l sin  x0  x02 
Now use the small angle approximation that cos   1   2 / 2 and sin    ,
x 
l 
2
2
 2l x0  x0 2 
  l 2 2 2l
 x0 1   2 
x0
  x0
 .
 

Expand the square root using the Taylor formula that for a  1 ,
 1  l 2 2 2l  
 l 2 2 2l 

x  x0 1  2 
  x0 1   2 
   x0  l .
x0
x0 
x0  

 2  x0
So we see that the spring has stretched by an amount x  l .
1  a   1 
a
hence
2