Question

In the figure, a meter stick swings about a pivot point at one end, at distance $h$ from the stick’s centre of mass. (a) What is the period of oscillation $T$ ? (b) What is the distance $L_{0}$ between the pivot point $O$ of the stick and the centre of oscillation of the stick? (c) If the physical pendulum of the figure $a$ and the associated sample problem is inverted and suspended at point $P$ , what is its period of oscillation? (d) Is the period now greater than, less than, or equal to its previous value? [Take $g = 9.8 m s^{- 2}$ ]

Accepted Answer

(a) The period for a physical pendulum is given by
$T = 2 π \sqrt{\frac{I}{m g h}}$ , for which we need the rotational inertia $g = 9.8 m s^{- 2}$ of the stick about the pivot point. We can treat the stick as a uniform rod of length $L$ and mass $m$ . Then moment of inertia $I = \frac{1}{3} m L^{2}$ , and the distance $h$ is $\frac{1}{2} L$ . Substituting these quantities in the above equation, we find
$T = 2 π \sqrt{\frac{I}{m g h}} = 2 π \sqrt{\frac{\frac{1}{3} m L^{2}}{m g (\frac{L}{2})}}$
$= 2 π \sqrt{\frac{2 L}{3 g}}$
$= 2 π \sqrt{\frac{2 \times 1}{3 \times 9.8}} = 1.64 s$
Note the result is independent of the pendulum’s mass $m$ .

(b) We want the length $L_{0}$ of the simple pendulum (drawn in figure $b$ ) that has the same period as the physical pendulum (the stick) of figure $a$ . $T = 2 π \sqrt{\frac{L_{0}}{g}} = 2 π \sqrt{\frac{2 L}{3 g}}$ .
You can see by inspection that
$L_{0} = \frac{2 L}{3}$
$\Rightarrow L_{0} = \frac{2}{3} \times 100 = 66.7 cm$
In figure $a$ , point $P$ marks this distance from suspension point $O$ . Thus, point $P$ is the stick’s centre of oscillation for the given suspension point. Point $P$ would be different for a different suspension choice.

(c)

Given,

The distance between $P$ and $C$ is, $h' = \frac{2}{3} L - \frac{1}{2} L = \frac{1}{6} L$ .

The length of the rod is, $L = 1 m$
After inversion, $P$ becomes the point of suspension.

The parallel axes theorem gives the moment of inertia,

$I = \frac{1}{12} m L^{2} + m {(h')}^{2} = (\frac{1}{12} + \frac{1}{36}) m L^{2} = \frac{1}{9} m L^{2}$

Now, time-period of physical pendulum is given as,

$T' = 2 π \sqrt{\frac{I}{m g (h')}} = 2 π \sqrt{\frac{L^{2} / 9}{g L / 6}} = 2 π \sqrt{\frac{2 L}{3 g}}$

Substitute the values in above equation, we get,

$\Rightarrow T' = 2 \times 3.14 \sqrt{\frac{2 \times 1}{3 \times 9.8}}$

$\Rightarrow T = 1.64 s$

(d)
As far as the characteristics of the periodic motion are concerned, the centre of oscillation provides a pivot that is equal to that of case $(a)$ .

Important Questions on Oscillations

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