Problem 1: The potential energy of an oscillating particle of mass “m” along straight line is given as :
U
x
=
a
+
b
x
−
c
2
U
x
=
a
+
b
x
−
c
2
The mechanical energy of the oscillating particle is “E”.
 Determine whether oscillation is SHM?
 If oscillation is SHM, then find amplitude and maximum kinetic energy.
Solution : If the motion is SHM, then restoring force is a conservative force. The potential energy is, then, defined such that :
ⅆ
U
=

F
ⅆ
x
ⅆ
U
=

F
ⅆ
x
⇒
F
=
−
ⅆ
U
ⅆ
x
=

2
b
x
−
c
⇒
F
=
−
ⅆ
U
ⅆ
x
=

2
b
x
−
c
In order to find the center of oscillation, we put F = 0.
F
=

2
b
x

c
=
0
⇒
x
−
c
=
0
⇒
x
=
c
F
=

2
b
x

c
=
0
⇒
x
−
c
=
0
⇒
x
=
c
This means that particle is oscillating about point x = c. The displacement of the particle in that case is “xc” – not “x”. This, in turn, means that force is proportional to negative of displacement, “xc”. Hence, particle is executing SHM.
Alternatively, put y = xc :
F
=

2
b
y
F
=

2
b
y
This means that particle is executing SHM about y = 0. This means xc = 0, which in turn, means that particle is executing SHM about x = c.
The mechanical energy is related to amplitude by the relation :
E
=
1
2
m
ω
2
A
2
E
=
1
2
m
ω
2
A
2
⇒
A
=
2
E
m
ω
2
⇒
A
=
2
E
m
ω
2
Now,
m
ω
2
=
k
=
2
b
m
ω
2
=
k
=
2
b
. Hence,
⇒
A
=
2
E
2
b
=
E
b
⇒
A
=
2
E
2
b
=
E
b
The potential energy is minimum at the center of oscillation i.e. when x = c. Putting this value in the expression of potential energy, we have :
⇒
U
min
=
a
+
b
c

c
2
=
a
⇒
U
min
=
a
+
b
c

c
2
=
a
It is important to note that minimum value of potential energy need not be zero. Now, kinetic energy is maximum, when potential energy is minimum. Hence,
K
max
=
E
−
U
min
=
E
−
a
K
max
=
E
−
U
min
=
E
−
a