# P Find the Time Period of Small Oscillations of the Following Systems. (A) a Metre Stick Suspended Through the 20 Cm Mark. - Physics

Sum

Find the time period of small oscillations of the following systems. (a) A metre stick suspended through the 20 cm mark. (b) A ring of mass m and radius r suspended through a point on its periphery. (c) A uniform square plate of edge a suspended through a corner. (d) A uniform disc of mass m and radius r suspended through a point r/2 away from the centre.

#### Solution

(a) Moment Of inertia $\left( I \right)$ about the point X is given by ,

I = IC.G + mh2

$= \frac{m l^2}{12} + m h^2$

$= \frac{m l^2}{12} + m \left( 0 . 3 \right)^2$

$= m\left( \frac{1}{12} + 0 . 09 \right)$

$= m\left( \frac{1 + 1 . 08}{12} \right)$

$= m\left( \frac{2 . 08}{12} \right)$

The time Period $\left( T \right)$ is given by,

$T = 2\pi\sqrt{\frac{I}{mgl}}$ $\text { where } I = \text{ the moment of inertia, and }$ $l= \text{ distance between the centre of gravity and the point of suspension . }$ $\text {On substituting the respective values in the above formula, we get: }$ $T = 2\pi\sqrt{\frac{2 . 08 m}{m \times 12 \times 9 . 8 \times 0 . 3}}$

$= 1 . 52 s$

(b) Moment Of inertia $\left( I \right)$ about A is given as,
I = IC.G. + mr2 = mr2 + mr2 = 2mr2

The time period (T) will be,

$T = 2\pi\sqrt{\frac{I}{mgl}}$

$\text { On substituting the respective values in the above equation, we have: }$

$T = 2\pi\sqrt{\frac{2m r^2}{mgr}}$

$= 2\pi\sqrt{\frac{2r}{g}}$

(c) Let I be the moment of inertia of a uniform square plate suspended through a corner.

$I = m\left( \frac{a^2 + a^2}{3} \right) = \frac{2m}{3} a^2$

In the

$\bigtriangleup$ ABC , l2 + l2 = a2

$\therefore l = \frac{a}{\sqrt{2}}$

$\Rightarrow T = 2\pi\sqrt{\frac{I}{mgl}}$

$= 2\pi\sqrt{\frac{2m a^2}{3mgl}}$

$= 2\pi\sqrt{\frac{2 a^2}{3ga\sqrt{2}}}$

$= 2\pi\sqrt{\frac{\sqrt{8}a}{3g}}$

(d)

$\text { We know }$

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

$\text { Distance between the C . G . and suspension point }, l = \frac{r}{2}$

Moment of inertia about A will be:
l = IC.G. + mh2

$= \frac{m r^2}{2} + m \left( \frac{r}{2} \right)^2$

$= m r^2 \left( \frac{1}{2} + \frac{1}{4} \right) = \frac{3}{4}m r^2$

Time period (T) will be,

$T = 2\pi\sqrt{\frac{I}{mgl}}$

$= 2\pi\sqrt{\frac{3m r^2}{4mgl}} = 2\pi\sqrt{\frac{3r}{2g}}$

Is there an error in this question or solution?
Chapter 12: Simple Harmonics Motion - Exercise [Page 255]

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 12 Simple Harmonics Motion
Exercise | Q 48 | Page 255

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