1.5 Simple and physical pendulum

Simple pendulum is an ideal oscillatory mechanism, which executes SHM. The restoring mechanism, in this case, is provided by gravitational force. Simple pendulum is simple in construction. It consists of a "particle" mass hanging from a string. The other end of the string is fixed. The overriding requirement of simple pendulum, executing SHM, can be stated in two supplementary ways :

• Mechanical energy of the oscillating system is conserved.
• The torque due to gravity (or angular acceleration) is proportional to negative of angular displacement.

Fulfillment of the requirements, as stated, imposes certain limitations on the construction and working of simple pendulum. It can be easily inferred that we probably can not fulfill the requirement stringently, but can only approximate at the best. In this module, therefore, we shall first analyze the motion for an ideal case and then deduce conditions, which need to be fulfilled to realize ideal case to the best possible extent.

Motion of simple pendulum

There are two forces acting on the particle mass hanging from the string (called pendulum bob). One is the gravity (mg), which acts vertically downward. Other is the tension (T) in the string. In equilibrium position, the bob hangs in vertical position with zero resultant force :

$$T-mg=0$$

At a displaced position, a net torque about the pivot point “O” acts on the pendulum bob which tends to restore its equilibrium position. In order to calculate net torque, we resolve gravity in two perpendicular components (i) mg cosθ along string and (ii) mg sinθ tangential to the path of motion.

Together with tension and two components of gravity, there are three forces acting on the pendulum bob. The line of action of tension and the component of gravity along string passes through pivot point, “O”. Therefore, torque about pivot point due to these two forces is zero. The torque on the pendulum bob is produced only by the tangential component of gravity. Hence, torque on the bob is :

$$\tau =\text{moment arm}X\text{Force}=-LXmg\sin \theta =-mgL\sin \theta$$

where "L" is the length of the string. We have introduced negative sign as torque is clockwise against the positive direction of displacement (anticlockwise). We can, now, use the relation “τ =Iα” to obtain the relation for angular acceleration :

$$\Rightarrow \tau =I\alpha =-mgL\sin \theta$$

$$\Rightarrow \alpha =-\frac{mgL}{I}\sin \theta$$

Clearly, this equation is not in the form “ $\alpha =-{\omega }^2\theta$ ” so that pendulum bob can execute SHM. It is evident that if the requirement of SHM is to be met, then

$$\sin \theta =\theta$$

Is it possible? Not exactly, but approximately yes - if the angular displacement is a small measure. Let us check out few values using calculator :

-------------------------------------------- Degree Radian sine value-------------------------------------------- 0 0.0 0.01 0.01746 0.017459 2 0.034921 0.0349143 0.052381 0.052357 4 0.069841 0.0697855 0.087302 0.087191 --------------------------------------------