We write various equations for angular SHM without derivation – unless there is differentiating aspect involved. In general, we substitute :
 Linear inertia “m” by angular inertia “I”
 Force, “F” by torque “τ”
 Linear acceleration “a” by angular acceleration, “α”
 Linear displacement “x” by angular displacement “θ”
 Linear amplitude “A” by angular amplitude “
θ
0
θ
0
”
 Linear velocity “v” by angular velocity “
ⅆ
θ
∕
ⅆ
t
ⅆ
θ
∕
ⅆ
t
”
Importantly, symbols of angular frequency (ω), spring constant (k), phase constant (φ), time period (T) and frequency (ν) remain same in the description of angular SHM.
Angular displacement
θ
=
θ
0
sin
ω
t
+
φ
θ
=
θ
0
sin
ω
t
+
φ
where “
θ
0
θ
0
” is the amplitude, “φ” is the phase constant and “ωt + φ” is the phase. Clearly, angular displacement is periodic with respect to time as it is represented by bounded trigonometric function. The displacement “θ” varies between “
−
θ
0
−
θ
0
” and “
θ
0
θ
0
”.
SHM Equation
ⅆ
2
θ
ⅆ
t
2
+
ω
2
θ
=
0
ⅆ
2
θ
ⅆ
t
2
+
ω
2
θ
=
0
Angular velocity
Angular velocity differs to angular frequency (ω). In the case of linear SHM, we had compared linear SHM with uniform circular motion (UCM). It was found that projection of UCM on an axis is equivalent description of linear SHM. It emerged that angular frequency is same as the magnitude of constant angular velocity of the equivalent UCM.
The angular velocity of the body under angular oscillation, however, is different. Importantly, angular velocity of SHM is not constant – whereas angular frequency is constant.
The angular velocity in angular SHM is obtained either as the solution of equation of motion or by differentiating expression of angular displacement with respect to time. Clearly, we need to have a different symbol to represent angular velocity in angular SHM. Let us denote this by the differential expression itself “
ⅆ
θ
∕
ⅆ
t
ⅆ
θ
∕
ⅆ
t
”, which is not equal to “ω”.
ⅆ
θ
ⅆ
t
=
ω
θ
0
2

θ
2
=
ω
θ
0
cos
ω
t
+
φ
ⅆ
θ
ⅆ
t
=
ω
θ
0
2

θ
2
=
ω
θ
0
cos
ω
t
+
φ
Angular acceleration
α
=

ω
2
θ
=

k
I
θ
=

ω
2
θ
0
sin
ω
t
+
φ
α
=

ω
2
θ
=

k
I
θ
=

ω
2
θ
0
sin
ω
t
+
φ
Torque
τ
=
I
α
=

k
θ
=

I
ω
2
θ
=
=

I
ω
2
θ
0
sin
ω
t
+
φ
τ
=
I
α
=

k
θ
=

I
ω
2
θ
=
=

I
ω
2
θ
0
sin
ω
t
+
φ
Frequency, angular frequency, time period
The angular frequency is given by :
ω
=
k
I
=

angular acceleration
angular displacement

ω
=
k
I
=

angular acceleration
angular displacement

Time period is obtained from the defining relation :
T
=
2
π
ω
=
2
π
I
k
=
2
π

angular displacement
angular acceleration

T
=
2
π
ω
=
2
π
I
k
=
2
π

angular displacement
angular acceleration

Frequency is obtained from the defining relation :
ν
=
1
T
=
ω
2
π
=
1
2
π
k
I
=
1
2
π

angular acceleration
angular displacement

ν
=
1
T
=
ω
2
π
=
1
2
π
k
I
=
1
2
π

angular acceleration
angular displacement

Kinetic energy
In terms of angular displacement :
K
=
1
2
I
d
θ
d
t
2
=
1
2
I
ω
2
θ
0
2
−
θ
2
K
=
1
2
I
d
θ
d
t
2
=
1
2
I
ω
2
θ
0
2
−
θ
2
In terms of time :
K
=
1
2
I
ω
2
θ
0
2
cos
2
ω
t
+
φ
K
=
1
2
I
ω
2
θ
0
2
cos
2
ω
t
+
φ
Potential energy
In terms of angular displacement :
U
=
1
2
k
θ
2
=
1
2
I
ω
2
θ
2
U
=
1
2
k
θ
2
=
1
2
I
ω
2
θ
2
In terms of time :
U
=
1
2
I
ω
2
θ
0
2
sin
2
ω
t
+
φ
U
=
1
2
I
ω
2
θ
0
2
sin
2
ω
t
+
φ
Mechanical energy
E
=
K
+
U
E
=
K
+
U
E
=
1
2
I
ω
2
θ
0
2
−
θ
2
+
1
2
I
ω
2
θ
2
=
1
2
I
ω
2
θ
0
2
E
=
1
2
I
ω
2
θ
0
2
−
θ
2
+
1
2
I
ω
2
θ
2
=
1
2
I
ω
2
θ
0
2
Problem 1: A body executing angular SHM has angular amplitude of 0.4 radian and time period of 0.1 s. If angular displacement of the body is 0.2 radian from the center of oscillation at time t = 0, then write the equation of angular displacement.
Solution : The equation of angular displacement is given by :
θ
=
θ
0
sin
ω
t
+
φ
θ
=
θ
0
sin
ω
t
+
φ
Here,
θ
0
=
0.4
radian
θ
0
=
0.4
radian
Also, time period is given. Hence, we can determine angular frequency, ω, as :
⇒
ω
=
2
π
T
=
2
π
0.1
=
20
π
radian/s
⇒
ω
=
2
π
T
=
2
π
0.1
=
20
π
radian/s
Putting these values, the expression of angular displacement is :
⇒
θ
=
0.4
sin
20
π
t
+
φ
⇒
θ
=
0.4
sin
20
π
t
+
φ
We, now, need to determine the phase constant for the given initial condition. At t = 0, θ = 0.2 radian. Hence,
⇒
0.2
=
0.4
sin
20
π
X
0
+
φ
=
0.4
sin
φ
⇒
0.2
=
0.4
sin
20
π
X
0
+
φ
=
0.4
sin
φ
⇒
sin
φ
=
0.2
0.4
=
1
2
=
sin
π
6
⇒
sin
φ
=
0.2
0.4
=
1
2
=
sin
π
6
⇒
φ
=
π
6
⇒
φ
=
π
6
Putting in the expression, the angular displacement is given by :
⇒
θ
=
0.4
sin
20
π
t
+
π
6
⇒
θ
=
0.4
sin
20
π
t
+
π
6