A horizontal rod of mass m and length L is pivoted smoothly at one end. The rod's other end is supported by a spring of force constant k as shown in figure. The rod is rotated (in vertical plane) by a small angle θ from its horizontal equilibrium position and released. The angular frequency of the subsequent simple harmonic motion is:

Question

A horizontal rod of mass m and length L is pivoted smoothly at one end. The rod's other end is supported by a spring of force constant k as shown in figure. The rod is rotated (in vertical plane) by a small angle θ from its horizontal equilibrium position and released. The angular frequency of the subsequent simple harmonic motion is: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/6f705c7d59885a88021f2883cb4b38de-.png" /> (A) √(3 k/m) (B) √(k/3 m) (C) √(3 k/m)+(3 g/2 L) (D) √(k/m)

Tardigrade Admin · Accepted Answer

Restoring torque on rod after small angle tilt is

τ_{R} = - k y L = - K L^{2} θ (Since y ≅ L θ from figure)

\Rightarrow k L^{2} θ = - \frac{m L ^{2}}{3} \cdot α

\Rightarrow α = - \frac{3 k}{m} \cdot θ

comparing with

α = - ω^{2} θ

We get

ω = \frac{3 k}{m}

Note: Torque due to

m g

was already balanced so it is not taken

$\frac{3 k}{m}$

$\frac{k}{3 m}$

$\frac{3 k}{m} + \frac{3 g}{2 L}$

$\frac{k}{m}$

m3k​​

3mk​​

m3k​+2L3g​​

mk​​

Restoring torque on rod after small angle tilt is τR​=−kyL=−KL2θ (Since y≅Lθ from figure ) ⇒kL2θ=−3mL2​⋅α ⇒α=−m3k​⋅θ comparing with α=−ω2θ We get ω=m3k​​ Note: Torque due to mg was already balanced so it is not taken

$\frac{3 k}{m}$

$\frac{k}{3 m}$

$\frac{3 k}{m} + \frac{3 g}{2 L}$

$\frac{k}{m}$