IDNLearn.com provides a platform for sharing and gaining valuable knowledge. Discover the information you need from our experienced professionals who provide accurate and reliable answers to all your questions.
Sagot :
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.
### 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.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com provides the answers you need. Thank you for visiting, and see you next time for more valuable insights.
