Question

A pendulum of length L and mass m has a spring of force constant k connected horizontally to it at a distance h below its point of suspension. The rod used for vertical suspension of length L is rigid and massless. The frequency of vibration of the system for small values of θ is

• A. $\frac{1}{2\pi L}\sqrt{gL + \frac{kh}{m}}$
• B. $\frac{1}{2\pi L}\sqrt{\frac{mgL + k}{m}}$
• C. 2π$\sqrt{\frac{mL^{2}}{mgL + kh}}$
• D. $\frac{1}{2\pi L}\sqrt{gL + \left( \frac{kh^{2}}{m} \right)}$

Answer 1

Answer: (D)

$\frac{1}{2\pi L}\sqrt{gL + \left( \frac{kh^{2}}{m} \right)}$

Answer 2

Torques of both spring and gravitational force starts acting as

restoring torque, after the pendulam has been displaced by

small angle θ.

Taking torques about O

-[mg (Lθ) + khθ (h)] = Iα= mL² α

α = - $\left( \frac{mgL + kh^{2}}{mL^{2}} \right)\theta = - \omega^{2}\theta$

ω = $\sqrt{\frac{mL^{2}}{mgL + kh^{2}}}$

f = $\frac{\omega}{2\pi} = \frac{1}{2\pi L}\sqrt{(gL) + \frac{kh^{2}}{m}}$