Question

A tall, thin tree sways back and forth in the breeze with a frequency of 2.90 Hz. What is the period of the tree?

Answer 1

The period of tree is 0.34 seconds

Explanation:

Given:

Frequency  of the tree, f = 2.9Hz

Time period, T = ?

We know,

$Time period = \frac{1}{frequency}$

$T = \frac{1}{2.9} \\\\T = 0.34 sec$

Therefore, the period of tree is 0.34 seconds

Answer 2

The time period of the swaying tall thin tree is 0.345 seconds.

Given the data in the question;

• frequency of the swaying tree; $f = 2.90Hz = 2.90s^{-1}$

• Period of the swaying tree; $T = \ ?$

Period and Frequency

Frequency is simply the number of occurrences of an event repeated per unit of time.

Period is simply the time taken for a complete cycle of vibration to pass a given point. Frequency is expressed as;

$f = \frac{1}{T}$

Where f is frequency and T is period of time.

To determine the period of the swaying, we substitute our given value into the expression above.

$f = \frac{1}{T} \\\\2.90s^{-1} = \frac{1}{T} \\\\T = \frac{1}{2.90s^{-1}}\\\\T = 0.345s$

Therefore, the time period of the swaying tall thin tree is 0.345 seconds.

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