In this type of SHM, the particle (or body) involved moves to and fro along an arc of a circle. Besides, the motion should be such that:
(a) It is oscillatory.
(b) The resultant torque acting on the particle about some fixed axis normal to the plane of motion, is proportional to its angular displacement from the mean position, and is always direction such that it has a tendency to bring the particle to its mean position.
Thus, if `tau` be the resultant torque, and `theta` be the angular displacement, in any position
then `tau prop ( - theta)`
` tau = -ktheta ........ (1)`
where `k` is constant of proportionality.
Now, if `alpha` be the angular acceleration and `I` the moment of inertia of the particle about the centre of rotation,
Then, `tau = I alpha`
`alpha= -(k/I)theta......................(2)`
`alphaprop(-theta)`
Equation (2) corresponds to equation (1) in linear SHM
`omega=sqrt(k/I)` or `T = (2 pi)/(omega) = 2 pisqrt(I/k)`
Substituting `alpha= (d^2 theta)/(dt^2)` and `k/I =omega^2` in eqn. (2)
Equation for angular SHM can be written as
`(d^2 theta)/dt^2 +omega^2theta =0`
Solving the above equation, equations for angular displacement and angular velocity can be obtained as
`theta = theta_0sin ( omega t + phi) ........ (3)`
and `omega = omega_0 cos ( omega t + phi ) ........ (4)`
and `omega = omega_0sqrt(theta^2-phi^2) ........ (5)`
where `theta` is the angular amplitude.
Notice that the symbol `omega` has been used to denote angular velocity of the particle.
NOTE
Equations corresponding to angular SHM can be obtained from equation pertaining to linear SHM, by replacing mass by moment of inertia, force by torque, linear displacement by angular displacement etc .
In this type of SHM, the particle (or body) involved moves to and fro along an arc of a circle. Besides, the motion should be such that:
(a) It is oscillatory.
(b) The resultant torque acting on the particle about some fixed axis normal to the plane of motion, is proportional to its angular displacement from the mean position, and is always direction such that it has a tendency to bring the particle to its mean position.
Thus, if `tau` be the resultant torque, and `theta` be the angular displacement, in any position
then `tau prop ( - theta)`
` tau = -ktheta ........ (1)`
where `k` is constant of proportionality.
Now, if `alpha` be the angular acceleration and `I` the moment of inertia of the particle about the centre of rotation,
Then, `tau = I alpha`
`alpha= -(k/I)theta......................(2)`
`alphaprop(-theta)`
Equation (2) corresponds to equation (1) in linear SHM
`omega=sqrt(k/I)` or `T = (2 pi)/(omega) = 2 pisqrt(I/k)`
Substituting `alpha= (d^2 theta)/(dt^2)` and `k/I =omega^2` in eqn. (2)
Equation for angular SHM can be written as
`(d^2 theta)/dt^2 +omega^2theta =0`
Solving the above equation, equations for angular displacement and angular velocity can be obtained as
`theta = theta_0sin ( omega t + phi) ........ (3)`
and `omega = omega_0 cos ( omega t + phi ) ........ (4)`
and `omega = omega_0sqrt(theta^2-phi^2) ........ (5)`
where `theta` is the angular amplitude.
Notice that the symbol `omega` has been used to denote angular velocity of the particle.
NOTE
Equations corresponding to angular SHM can be obtained from equation pertaining to linear SHM, by replacing mass by moment of inertia, force by torque, linear displacement by angular displacement etc .