To find the frequency of the simple harmonic motion, we need to analyze the given expression for displacement and relate it to the parameters of simple harmonic motion.

Given information:

• Mass, $m = 2.00 \, \text{kg}$
• Spring constant, $k = 900 \, \text{N/m}$
• Displacement equation: $x(t) = 12.0 \, \text{cm} \sin(\omega t)$

Step-by-Step Solution:

1. Convert the displacement to meters: $x(t) = 0.12 \, \text{m} \sin(\omega t)$

2. Determine the angular frequency, $\omega$: For a spring-mass system in simple harmonic motion, the angular frequency $\omega$ is given by: $\omega = \sqrt{\frac{k}{m}}$

3. Calculate $\omega$: $\omega = \sqrt{\frac{900}{2.00}} = \sqrt{450} \approx 21.21 \, \text{rad/s}$

4. Convert $\omega$ to frequency $f$: The frequency $f$ is related to the angular frequency $\omega$ by: $f = \frac{\omega}{2 \pi}$ Substituting $\omega = 21.21$: $f = \frac{21.21}{2 \pi} \approx 3.38 \, \text{Hz}$

Conclusion:

The correct answer is: 3.38 Hz

Let me know if you want further clarification on any part of the solution!

Here are 5 questions to extend understanding:

1. How does mass affect the frequency of simple harmonic motion?
2. What would happen to the frequency if the spring constant doubled?
3. How would you calculate the period of this motion?
4. What role does amplitude play in simple harmonic motion?
5. Can you determine the total energy of the system using given parameters?

Tip: For spring-mass systems, increasing mass will decrease the frequency of oscillation because it slows down the system’s response.