First slide

Applications of SHM

Question

A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle θ in one direction and released. The frequency of oscillation is

Moderate

A

$\frac{1}{2 π} \sqrt{\frac{2 k}{M}}$
B

$\frac{1}{2 π} \sqrt{\frac{k}{M}}$
C

$\frac{1}{2 π} \sqrt{\frac{6 k}{M}}$
D

$\frac{1}{2 π} \sqrt{\frac{24 k}{M}}$

Solution

The rod will execute angular oscillations
Torque method:
Let rod be turned by small angle $θ$ .
Extension in each of the springs = $\frac{L θ}{2}$
Force applied by each spring = $\frac{k L θ}{2}$
Torque on the system = $- 2 (\frac{k L θ}{2}) (\frac{L}{2}) = \frac{k L^{2} θ}{2}$
$I α = - \frac{k L^{2} θ}{2}$
$\Rightarrow \frac{M L^{2}}{12} α = - \frac{k L^{2} θ}{2}$
$α = - \frac{6 k}{M} θ$
Angular acceleration is proportional, and directed opposite, to angular displacement. Hence SHM.
$\Rightarrow f = \frac{1}{2 π} \sqrt{\frac{α}{θ}} = \frac{1}{2 π} \sqrt{\frac{6 k}{M}}$
Energy method:
Let $θ$ be angular displacement at a given time, and $ω_{0}$ be the angular velocity of the rod at that time.
So, $ω_{0} = \frac{d θ}{d t} a n d α = \frac{d ω_{0}}{d t}$
Energy of the oscillating system
$E = \frac{1}{2} k x^{2} + \frac{1}{2} k x^{2} + \frac{1}{2} I ω_{0}^{2}$ , where x is the extension in each spring.
But for small extension, $x = \frac{L θ}{2}$
So, $E = 2 \frac{1}{2} k {(\frac{L θ}{2})}^{2} + \frac{1}{2} (\frac{M L^{2}}{12}) ω_{0}^{2}$
Energy is constant. So, $\frac{d E}{d t} = 0$
$\Rightarrow k \frac{L^{2}}{4} 2 θ (\frac{d θ}{d t}) + \frac{1}{2} (\frac{M L^{2}}{12}) 2 ω_{0} (\frac{d ω_{0}}{d t}) = 0$ $\Rightarrow \frac{k}{2} θ + (\frac{M}{12}) α = 0$
$\Rightarrow α = - \frac{6 k}{M} θ$
Angular acceleration is proportional, and directed opposite, to angular displacement. Hence SHM.
$\Rightarrow f = \frac{1}{2 π} \sqrt{\frac{α}{θ}} = \frac{1}{2 π} \sqrt{\frac{6 k}{M}}$

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