Let's approach this problem step by step.
### Given Data:
- Mass [tex]\(m = 8\)[/tex] kg
- Amplitude [tex]\(A = 30\)[/tex] cm = 0.30 m (conversion to meters)
- Displacement [tex]\(x = 30\)[/tex] cm = 0.30 m (conversion to meters initially given)
- Restoring force at this displacement [tex]\(F = 60\)[/tex] N
- New displacement for the energy calculations [tex]\(x' = 2\)[/tex] cm = 0.02 m (conversion to meters)
### Step-by-Step Solution:
#### a. Finding the Period (T):
1. Hooke's Law to Find Spring Constant (k):
The restoring force [tex]\(F\)[/tex] is related to the spring constant [tex]\(k\)[/tex] and displacement [tex]\(x\)[/tex] by Hooke's Law: [tex]\( F = kx \)[/tex].
[tex]\[ k = \frac{F}{x} = \frac{60 \, \text{N}}{0.30 \, \text{m}} = 200 \, \text{N/m} \][/tex]
2. Angular Frequency (ω):
The angular frequency for a mass-spring system is given by:
[tex]\[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200 \, \text{N/m}}{8 \, \text{kg}}} = 5 \, \text{rad/s} \][/tex]
3. Period (T):
The period [tex]\(T\)[/tex] is related to the angular frequency by:
[tex]\[ T = \frac{2\pi}{\omega} \][/tex]
[tex]\[ T = \frac{2\pi}{5 \, \text{rad/s}} \approx 1.2566 \, \text{seconds} \][/tex]
#### b. Finding the Acceleration at x' = 2 cm:
The acceleration in a simple harmonic motion (SHM) is given by:
[tex]\[ a = -\omega^2 x' \][/tex]
Substituting the given values:
[tex]\[ a = -(5 \, \text{rad/s})^2 \cdot 0.02 \, \text{m} \][/tex]
[tex]\[ a = -0.5 \, \text{m/s}^2 \][/tex]
#### c. Finding the Potential and Kinetic Energy at x' = 2 cm:
1. Potential Energy (U):
The potential energy in a spring system is given by:
[tex]\[ U = \frac{1}{2}kx'^2 \][/tex]
Substituting the values:
[tex]\[ U = \frac{1}{2} \cdot 200 \, \text{N/m} \cdot (0.02 \, \text{m})^2 \][/tex]
[tex]\[ U = 0.04 \, \text{J} \][/tex]
2. Total Mechanical Energy (E):
The total mechanical energy in SHM is conserved and is given by the potential energy at maximum displacement (amplitude):
[tex]\[ E = \frac{1}{2}kA^2 \][/tex]
[tex]\[ E = \frac{1}{2} \cdot 200 \, \text{N/m} \cdot (0.30 \, \text{m})^2 \][/tex]
[tex]\[ E = 9 \, \text{J} \][/tex]
3. Kinetic Energy (K) at x' = 2 cm:
The kinetic energy is the total mechanical energy minus the potential energy at [tex]\(x'\)[/tex]:
[tex]\[ K = E - U \][/tex]
[tex]\[ K = 9 \, \text{J} - 0.04 \, \text{J} \][/tex]
[tex]\[ K = 8.96 \, \text{J} \][/tex]
### Summary:
- Period (T): [tex]\(1.2566\)[/tex] seconds
- Acceleration (a) at 2 cm: [tex]\(-0.5 \)[/tex] m/s²
- Potential Energy (U) at 2 cm: [tex]\(0.04\)[/tex] Joules
- Kinetic Energy (K) at 2 cm: [tex]\(8.96\)[/tex] Joules
With this step-by-step solution, all parts of the given problem are addressed and the results are consistent with the problem’s constraints.
