We write restoring torque equation for angular SHM as :

τ = - k θ τ = - k θ

This “k” is springiness of the restoring torque. We associate “springiness” with any force or torque which follows the linear proportionality with negative displacement. For this reason, we call it “spring constant” for all system – not limited to block-spring system. In the case of simple pendulum, we associate this “springiness” to gravity. Similarly, this property can be associated with other forces like torsion, stress, pressure and many other force systems, which operate to restore equilibrium.

We write various equations for angular SHM without derivation – unless there is differentiating aspect involved. In general, we substitute :

1. Linear inertia “m” by angular inertia “I”
2. Force, “F” by torque “τ”
3. Linear acceleration “a” by angular acceleration, “α”
4. Linear displacement “x” by angular displacement “θ”
5. Linear amplitude “A” by angular amplitude “ θ 0 θ 0 ”
6. 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

Angular SHM provides a very effective technique for measuring moment of inertia. We can have a set up of torsion pendulum for a body, whose MI is to be determined. Measuring time period of oscillation, we have :

T = 2 π I k T = 2 π I k

Squaring both sides and arranging,

⇒ k T 2 = 4 π 2 I ⇒ k T 2 = 4 π 2 I

⇒ I = k T 2 4 π 2 ⇒ I = k T 2 4 π 2

Generally, we do not use this equation in the present form as it involves unknown spring constant, “k”. Actually, we carry out experiment for measuring time period first with a regularly shaped object like a rod of known dimensions and mass. This allows us to determine spring constant as MI of the object is known. Then, we determine time period with the object whose MI is to be determined.

Let subscripts “1” and “2” denote objects of known and unknown MIs respectively. Then,

⇒ I 1 I 2 = T 1 2 T 2 2 ⇒ I 1 I 2 = T 1 2 T 2 2

⇒ I 2 = I 1 T 2 2 T 1 2 ⇒ I 2 = I 1 T 2 2 T 1 2