1.5 Simple and physical pendulum  (Page 3/3)

Frame of reference is accelerating upwards

Let us consider downward direction as positive. Referring left picture in the figure,

$$\Rightarrow g\prime =g-\left(-a\right)=g+a$$

$$\Rightarrow T=2\pi \sqrt{\left(\frac{L}{g\prime }\right)}=2\pi \sqrt{\left(\frac{L}{g+a}\right)}$$

Frame of reference is accelerating downwards

Let us consider downward direction as positive. Referring right picture in the figure above,

$$\Rightarrow g\prime =g-a$$

$$\Rightarrow T=2\pi \sqrt{\left(\frac{L}{g\prime }\right)}=2\pi \sqrt{\left(\frac{L}{g-a}\right)}$$

If the simple pendulum is falling freely, then a = g,

$$\Rightarrow T=2\pi \sqrt{\left(\frac{L}{g-g}\right)}=\text{undefined}$$

Thus, free falling pendulum will not oscillate.

Frame of reference is moving in horizontal direction

Horizontal direction is perpendicular to the vertical direction of gravity. Hence, magnitude of effective acceleration is given as :

$$\Rightarrow g\prime =\sqrt{\left(g^2+a^2\right)}$$

$$\Rightarrow T=2\pi \sqrt{\left(\frac{L}{g\prime }\right)}=2\pi \sqrt{\left(\frac{L}{\sqrt{\left(g^2+a^2\right)}}\right)}$$

Change in length

The length of simple can change due to change in temperature. The change in length is given by :

$$L\prime =L\left(1+\alpha \Delta \theta \right)$$

Where “α” and “θ” are temperature coefficient and temperature respectively.

We should be careful in using this relation in case when simple pendulum is driver of a clock. For example, an increase in temperature translates into increase in length, which in turn translates into an increase in time period. However, an increase in time period translates into loss of time on the scale of the watch. In the nutshell, watch will run slow.

Physical pendulum

In this case, a rigid body – instead of point mass - is pivoted to oscillate as shown in the figure. There is no requirement of string. As a result, there is no tension involved in this case. Besides these physical ramifications, the working of compound pendulum is essentially same as that of simple pendulum except in two important aspects :

• Gravity acts through center of mass of the rigid body. Hence, length of pendulum used in equation is equal to linear distance between pivot and center of mass (“h”).
• The moment of inertia of the rigid body about point suspension is not equal to " $m{L}^2$ " as in the case of simple pendulum.

The time period of compound pendulum, therefore, is given by :

$$T=\frac{2\pi }{\omega }=2\pi \sqrt{\left(\frac{I}{mgh}\right)}$$

In case we know MI of the rigid body, we can evaluate above expression of time period for the physical pendulum. For illustration, let us consider a uniform rigid rod, pivoted from a frame as shown in the figure. Clearly, center of mass is at a distance “L/2” from the point of suspension :

$$h=\frac{L}{2}$$

Now, MI of the rigid rod about its center is :

$$I_C=\frac{m{L}^2}{12}$$

We are , however, required to evaluate MI of the rod about the point of suspension, i.e. “O”. Applying parallel axes theorem,

$$\Rightarrow I_O=\frac{m{L}^2}{12}+m{\left(\frac{L}{2}\right)}^2=\frac{m{L}^2}{3}$$

Putting in the equation of time period, we have :

$$\Rightarrow T=2\pi \sqrt{\left(\frac{I}{mgh}\right)}=2\pi \sqrt{\left(\frac{2m{L}^2}{3mgL}\right)}=2\pi \sqrt{\left(\frac{2L}{3g}\right)}$$

The important thing to note about this relation is that time period is still independent of mass of the rigid body. However, time period is not independent of mass distribution of the rigid body. A change in shape or size or change in mass distribution will change MI of the rigid body about point of suspension. This, in turn, will change time period.

Further, we should note that physical pendulum is an effective device to measure “g”. As a matter of fact, this device is used extensively in gravity surveys around the world. We only need to determine time period or frequency to determine the value of “g”. Squaring and rearranging,

$$\Rightarrow g=\frac{8{\pi }^{2}L}{3T^2}$$

Point of oscillation

We can think of physical pendulum as if it were a simple pendulum. For this, we can consider the mass of the rigid body to be concentrated at a single point as in the case of simple pendulum such that time periods of two pendulums are same. Let this point be at a linear distance " $L_o$ " from the point of suspension. Here,

$$\Rightarrow T=2\pi \sqrt{\left(\frac{I}{mgh}\right)}==2\pi \sqrt{\left(\frac{L_0}{g}\right)}$$

$$\Rightarrow Lo=\frac{Ig}{mgh}$$

The point defined by the vertical distance, " $L_o$ ", from the point of suspension is called point of oscillation of the physical pendulum. Clearly, point of oscillation will change if point of suspension is changed.