Question

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 107 m. The relative change in the angular frequency of the pendulum is best given

Answer 1

0.2675

Answer 2

1. We first found the gravitational acceleration g using the initial pendulum conditions: $\omega_0 = \sqrt{\frac{g}{l}} \Rightarrow g = \omega_0^2l = 100$ m/s²

2. Then we used the formula for relative change in frequency due to parametric oscillation: $\frac{\Delta \omega}{\omega_0} = \frac{a\Omega^2}{4g}$

3. Substituting the values: $\frac{\Delta \omega}{\omega_0} = \frac{107 \times 1^2}{4 \times 100} = 0.2675$

This result shows that the vertical oscillation of the support causes a significant change in the pendulum's angular frequency, increasing it by approximately 27% of its original value.