First slide

Applications of SHM

Question

A physical pendulum pivoted at a point executes angular oscillations. Its mass m,has its centre of mass at distance r from the point of suspension. If its moment of inertia is I, then its angular frequency is

Moderate

A

$ω = \sqrt{\frac{M g r}{I^{2}}}$
B

$ω = \sqrt{\frac{I^{2}}{M g r}}$
C

$ω = \sqrt{\frac{M g r}{I}}$
D

$ω = \sqrt{\frac{I}{M g r}}$

Solution

For a body executing simple harmonic motion, Restoring torque acting on it after a small displacement $θ$ , about an axis, $τ = - c θ$
We know that this results in angular oscillations which can be also related $τ = I α$
by the equation
Equating the above two we get $- c θ = I α$
Where for angular oscillations $α = - ω^{2} θ$
Hence $- c θ = - I ω^{2} θ$
Simplifying this $c = I ω^{2}$
Hence $ω = \sqrt{\frac{c}{I}}$
For angular oscillation $ω = \sqrt{\frac{c}{I}}$ , where $ω = \frac{2 π}{T}$
Restoring torque acting on a rod after a small displacement $θ$ , about an axis passing through point of contact O with the curved path,
$τ = - M g r \sin θ$ , for small angles $\sin θ \approx θ$
Therefore, $τ = - c θ = - M g r θ$
Hence $c = M g r$ . Substituting this is (1) we get
$ω = \sqrt{\frac{M g r}{I}}$ .

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