Question

Chris pulls a slinky a distance of 24 inches from its equilibrium position and then releases it. The time for one oscillation is 3 seconds.

Answer 1

The frequency of Chris's slinky oscillation, pulled 24 inches and released, is 0.33 Hz. This means it completes approximately one-third of an oscillation in one second.

The frequency of oscillation for Chris's slinky can be calculated using the formula:

`\[ \text{Frequency} (f) = \frac{1}{\text{Time period} (T)} \]`

In the given statement, it is mentioned that the time for one complete oscillation (time period,
`\(T\))` is 3 seconds. Therefore, we can substitute this value into the formula:

`\[ f = \frac{1}{3 \, \text{seconds}} \]`

This implies that the slinky undergoes one complete oscillation every 3 seconds. The frequency, in this context, represents the number of oscillations per unit time. In this case, it is the reciprocal of the time period.

So, the frequency of Chris's slinky oscillation is
`\( (1)/(3) \)` oscillation per second or 0.33 Hz (Hertz).

In summary, Chris's slinky has a frequency of 0.33 Hz, meaning it completes approximately one-third of an oscillation in one second. This provides a quantitative measure of the slinky's oscillatory behavior based on the information provided in the statement.

The probable question maybe:

What is the frequency of oscillation for Chris's slinky as it undergoes harmonic motion, given that he pulls it 24 inches from its equilibrium position and the time for one complete oscillation is 3 seconds?