Name: Superposition of Two or More SHMs
Uploaded: 2019-07-19
Duration: 9 min 11 s
Description: What happens when two or more simple harmonic motions are superimposed? Watch this video to learn what would be the result of the superposition of SHMs.

Question

Compound pendulums are made of :

(a) A rod of length $l$ suspended through a point located at distance $l / 4$ from centre of rod.

(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.

Find the time period in each case.

Accepted Answer

Time period compound pendulum is given as $T = 2 π \sqrt{\frac{I}{m g h}}$ , where $I$ is moment of inertia of the body about axis of rotation, $m$ is mass and $h$ is centre of distance of axis of rotation from centre of mass.

$(a)$

Moment of inertia of rod about point $O$ is given as $I_{O} = I_{COM} + m {(\frac{l}{4})}^{2} \Rightarrow I_{O} = \frac{m l^{2}}{12} + \frac{m l^{2}}{16} = \frac{m l^{2}}{12} (1 + \frac{3}{4}) = \frac{7 m l^{2}}{48}$ .

Hence, time period is $T = 2 π \sqrt{\frac{(\frac{7 m l^{2}}{48})}{m g (\frac{l}{4})}} = 2 π \sqrt{\frac{7 m l^{2} (4)}{48 m g l}} = 2 π \sqrt{\frac{7 l}{12 g}} = π \sqrt{\frac{7 l}{3 g}}$ .

$(b)$

Moment of inertia of ring about point $O$ is given as

$I_{O} = I_{COM} + m {(r)}^{2} \Rightarrow I_{O} = m r^{2} + m r^{2} = 2 m r^{2}$ .

Hence, time period is $T = 2 π \sqrt{\frac{2 m r^{2}}{m g (r)}} = 2 π \sqrt{\frac{2 r}{g}}$ .

$(c)$

Moment of inertia of square about point $O$ is given as

$I_{O} = I_{COM} + m {(\frac{a}{\sqrt{2}})}^{2} \Rightarrow I_{O} = \frac{m a^{2}}{6} + \frac{m a^{2}}{2} = \frac{2}{3} m a^{2}$ .

Hence, time period is $T = 2 π \sqrt{\frac{\frac{2}{3} m a^{2}}{m g (\frac{a}{\sqrt{2}})}} = 2 π \sqrt{\frac{2 \sqrt{2} a}{3 g}}$ .

$(d)$

Moment of inertia of disc about point $O$ is given as

$I_{O} = I_{COM} + m {(r)}^{2} = \frac{m r^{2}}{2} + m r^{2} = \frac{3}{2} m r^{2}$ .

Hence, time period is $T = 2 π \sqrt{\frac{\frac{3}{2} m r^{2}}{m g (r)}} = 2 π \sqrt{\frac{3 r}{2 g}}$ .

Important Questions on Simple Harmonic Motion

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