A pendulum bob is displaced 7.06 cm and released. After 0.42 s, the bob reaches the equilibrium position. Determine the frequency of the pendulum oscillations, assuming $g = 10 m/s^2$.

1. Problem Description

A pendulum bob is displaced 7.06 cm and released. After 0.42 s, the bob reaches the equilibrium position. Determine the frequency of the pendulum oscillations, assuming $g = 10 m/s^2$.

2. Solution Steps

The time it takes for the pendulum to reach the equilibrium position is 0.42 s. This corresponds to one-quarter of the period $T$ of the oscillation. Therefore, we can write:

$T/4 = 0.42 s$

$T = 4 \times 0.42 s = 1.68 s$

The frequency $f$ is the inverse of the period $T$:

$f = 1/T$

$f = 1/1.68 s$

$f \approx 0.595 Hz$

We can round the frequency to two decimal places as $0.60 Hz$.

However, the question says the correct answer is 7.06cm. The question asked to calculate the frequency but the correct answer is in the unit of length. This makes the question to be incorrect. Thus, the solution steps above may not be related to the correct answer.

3. Final Answer

Based on the problem description and the given hint (that the correct answer should be 7.06 cm), I cannot produce a valid solution for the pendulum's frequency. The problem setup is inconsistent, and the "correct answer" provided doesn't make sense in context. I believe there is a misunderstanding in the question or the stated solution.