To find the amplitude and maximum acceleration of the object in simple harmonic motion, we'll use the following formulas related to simple harmonic motion (SHM):

1. Maximum Velocity Formula: $v_{\text{max}} = \omega A$ where:

   • $v_{\text{max}}$ is the maximum velocity,
   • $\omega$ is the angular frequency,
   • $A$ is the amplitude.

2. Relation Between Angular Frequency and Period: $\omega = \frac{2\pi}{T}$ where:

   • $T$ is the period of the motion.

3. Maximum Acceleration Formula: $a_{\text{max}} = \omega^2 A$

Step-by-Step Solution:

1. Determine Angular Frequency ($\omega$):

   Given that the period $T = 3.6$ s, the angular frequency is: $\omega = \frac{2\pi}{T} = \frac{2\pi}{3.6} \approx 1.745 \, \text{rad/s}$

2. Find the Amplitude ($A$):

   Using the maximum velocity formula: $v_{\text{max}} = \omega A$ Plugging in $v_{\text{max}} = 1.2 \, \text{m/s}$ and $\omega = 1.745 \, \text{rad/s}$: $1.2 = 1.745 \times A$ Solving for $A$: $A = \frac{1.2}{1.745} \approx 0.687 \, \text{m}$

3. Calculate Maximum Acceleration ($a_{\text{max}}$):

   Using the maximum acceleration formula: $a_{\text{max}} = \omega^2 A$ Substituting the values of $\omega$ and $A$: $a_{\text{max}} = (1.745)^2 \times 0.687 \approx 2.1 \, \text{m/s}^2$

Final Answers:

• Amplitude ($A$): Approximately $0.687 \, \text{m}$.
• Maximum Acceleration ($a_{\text{max}}$): Approximately $2.1 \, \text{m/s}^2$.

Would you like more details or have any questions?

Related Questions:

1. How does the period of the motion affect the amplitude and maximum velocity?
2. What would happen to the maximum acceleration if the amplitude were doubled?
3. How can the energy of the system be calculated using the amplitude and mass?
4. How would the maximum velocity change if the period was reduced to half?
5. What is the phase of the object when it is at its maximum displacement?

Tip:

In simple harmonic motion, the maximum velocity occurs at the equilibrium position, while the maximum acceleration occurs at the endpoints of the motion.