Question

A simple harmonic oscillator with an amplitude of 3.0 cm passes through its equilibrium position once every 2.0 seconds, what is the frequency of the oscillator?

a. 0.33 Hz
b. 2.0 Hz
c. 3.0 Hz
d. 0.25 Hz
e. 0.50 Hz
f. 1.5 Hz

Answer 1

The frequency of a simple harmonic oscillator passing through its equilibrium position once every 2.0 seconds is e. 0.50 Hz.

Step-by-step explanation:

The question asks for the frequency of a simple harmonic oscillator that passes through its equilibrium position once every 2.0 seconds. The frequency is the reciprocal of the period, which is the time taken for one complete cycle.

In this case, the period (T) is 2.0 seconds. Therefore, the frequency (f) can be calculated as

f = 1/T.

Using this formula:

• f = 1/T
• f = 1/2.0 s
• f = 0.50 Hz

Hence, the correct answer is e. 0.50 Hz.