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

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प्रश्न

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.

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उत्तर

(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,

\[T = 2\pi\sqrt{\frac{I}{mgl}}\] 

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

 

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पाठ 12: Simple Harmonics Motion - Exercise [पृष्ठ २५५]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 12 Simple Harmonics Motion
Exercise | Q 48 | पृष्ठ २५५

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