Download Document

Differential Equation of the
Mechanical Oscillator
Prepared by;
Dr. Rajesh Sharma
Assistant Professor
Dept of Physics
P.G.G.C-11, Chandigarh
Email: [email protected]
• The Hook’s law
F  Sx
(1)
• Let there be a body of mass m attached to a spring. Then according
to the Newton’s Second law 2of motion, we have
d x
F m 2
dt
• Substituting this in eq.(1), we get
d 2x
m 2   Sx
dt
d 2x
S


x
Or,
(2)
2
dt
m
S
2
• Putting
 w0 where w0 is another constant for the oscillator, we
m
get
d 2x
2


w
0x
2
dt
Or,
(3)
d 2x
2
 w0 x  0
2
dt
This is called the Differential Eq. of SHM and its solution is of the form
x  A cos w0t
Where A is the amplitude and w0 = S/m; T=2 m/s
Energy of the Mechanical Oscillator
• The particle executing SHM possesses Kinetic as well as
Potential energies.
KINETIC ENERGY:
• The KE of the body of mass m, when possessing the velocity
v=dx/dt is given by
2
1 2 1  dx 
U k  mv  m 
(4)
2
2  dt 
dx
v

  Aw0 sin w0t
• Since x  A cos w0t
and
dt
• 
Or,
Uk 
1
2
m Aw0 sin w0t 
2
Uk 
1
mA2w02 sin 2 w0t
2
(5)
POTENTIAL ENERGY:
• The particle executing SHM is moved against the restoring force and the
work so done is stored as the potential energy. Let us displace the particle
by dx , then the work done which is equal to the potential energy stored in
the system is given by:
dU p   Fdx
• Here the –ve sign indicates the work done against the restoring force.
Now, for the mass attached to the spring, we have F  Sx .
dU p   Sx dx  Sxdx
• We have
• The potential energy of the oscillator, when the displacement is x,
becomes
U  dU  Sxdx
p

p

1 2
U p  Sx
• Or,
2
• Putting x  A cos w0t we get
1 2
U p  SA cos 2 w0t
2
(6)
(7)
Total energy of the oscillator
• The total energy of the oscillator is given by
Um  Uk U p
1
1
• Or,
U m  mA2w02 sin 2 w0t  SA2 cos 2 w0t
2
2
S
 w02 therefore, S  mw02 . Hence,
• But,
m
1
1
U m  mA2w02 sin 2 w0t  mA2w02 cos 2 w0t
2
2
• Or,
1
U m  mA2w02 sin 2 w0t  cos 2 w0t 
2
1
• Or,
U m  mA2w02
2
(8)
• This shows that the total energy of the mechanical oscillator is constant or
is conserved and is independent of the location of the particle. I depends
upon m, A, w0 or S (because w0 = S/m).
Average Kinetic Energy
• From Eq.(5), we have
1
U k  mA2w02 sin 2 w0t
2
• The Average KE over the time period is given by
T

1 1
2 2
2
U k   mA w0  sin w0tdt 
T 2
0

T
1 1
 1  cos 2w0t 
  mA2w02  
dt
T 2
2
0
T
T


1 1
1
1

2 2
  mA w0    dt   cos 2w0tdt 
T 2
20
 2 0

T
1

T
• Therefore
1 
1
2 2 
 2 mA w0   t   0

 2
 
0
1 1
1
2 2  1 
U k   mA w0   T   mA2w02
T 2
 2  4
 T

  cos 2w0tdt  0
 0

Average Potential Energy
• The potential energy
1 2
U p  SA cos 2 w0t
2
• Putting S  mw02 we get
Up 
1
mA2w02 cos 2 w0t
2
• The Average potential energy overT the time period is given by

1 1
2 2
2
U p   mA w0  cos w0tdt 
T 2
0

T
1 1
 1  cos 2w0t 
  mA2w02  
dt
T 2
2
0
T
T


1 1
1
1

2 2
  mA w0    dt   cos 2w0tdt 
T 2
20
 2 0

T
1

T
1
Up 
T
1 
1
2 2 
 2 mA w0   t   0

 2
 
0
1
1
2 2 1 
2 2
mA
w
T

mA
w0
0 
 2

 2  4
 T

  cos 2w0tdt  0
 0

• Also, the total energy of the oscillator is
Um 
• Which shows that
1
mA2w02
2
1
U p  Uk  Um
2
• That is, the average potential energy is equal to the average kinetic energy
over a period and is equal to half the total energy of the mechanical
oscillator.