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 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

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 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 (A) (1/2 π) √ (2k/M) (B) (1/2 π) √ (k/M) (C) (1/2 π) √ (6k/M) (D) (1/2 π) √ (24k/M)

Tardigrade Admin · Accepted Answer

x = (L/2) θ Restoring torque = - (2kx) (L/2) propto = - ( kL(L/ 2 θ )/I) = - [ ( k L2/2/ ML2 /12) ] θ = - ( (6 k/M) ) θ ∴ f =(1/2 π) √ |( propto/ θ )| =( 1/2π) √( 6k/M)

$\frac{1}{2 π} \frac{2 k}{M}$

$\frac{1}{2 π} \frac{k}{M}$

$\frac{1}{2 π} \frac{6 k}{M}$

$\frac{1}{2 π} \frac{24 k}{M}$

$x = \frac{L}{2} θ$
Restoring torque = $- (2 k x) \frac{L}{2}$
$\propto= - \frac{k L ( L /2 θ )}{I} = - [\frac{k L ^{2} /2}{M L ^{2} /12}] θ = - (\frac{6 k}{M}) θ$
$∴ f = \frac{1}{2 π} ∣ \frac{\propto}{θ} ∣ = \frac{1}{2 π} \frac{6 k}{M}$

2π1​M2k​​

2π1​Mk​​

2π1​M6k​​

2π1​M24k​​

x=2L​θ Restoring torque = −(2kx)2L​ ∝=−IkL(L/2θ)​=−[ML2/12kL2/2​]θ=−(M6k​)θ ∴f=2π1​∣θ∝​∣​=2π1​M6k​​

$\frac{1}{2 π} \frac{2 k}{M}$

$\frac{1}{2 π} \frac{k}{M}$

$\frac{1}{2 π} \frac{6 k}{M}$

$\frac{1}{2 π} \frac{24 k}{M}$

$x = \frac{L}{2} θ$
Restoring torque = $- (2 k x) \frac{L}{2}$
$\propto= - \frac{k L ( L /2 θ )}{I} = - [\frac{k L ^{2} /2}{M L ^{2} /12}] θ = - (\frac{6 k}{M}) θ$
$∴ f = \frac{1}{2 π} ∣ \frac{\propto}{θ} ∣ = \frac{1}{2 π} \frac{6 k}{M}$