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Sagot :
Certainly! Let's break down each term and concept, then solve the given problem step-by-step.
### Definitions
#### Amplitude:
Amplitude is the maximum displacement of a particle from its equilibrium position in simple harmonic motion (SHM).
#### Period:
The period is the time taken for one complete cycle of motion.
#### Simple Pendulum:
A simple pendulum consists of a mass (often called the bob) attached to a string or rod of fixed length, swinging back and forth under the influence of gravity.
### Examples of Simple Harmonic Motion (SHM):
1. A mass-spring system, where a mass is attached to a spring and can oscillate back and forth.
2. A simple pendulum, where a mass is attached to a string and swings back and forth.
### Problem Solution:
Let's find the various quantities step-by-step:
#### Given Data:
1. Mass of the particle (m): 0.30 kg
2. Amplitude (A): 0.04 m
3. Period (T): 21 s
4. Small bob mass: 0.20 kg (or [tex]\(20 \times 10^{-2}\)[/tex] kg)
5. Small bob length: 0.02 m (or [tex]\(2 \times 10^{-2}\)[/tex] m)
6. Small bob period: 21 s
7. Small bob amplitude: 0.02 m (or [tex]\(2 \times 10^{-2}\)[/tex] m)
We need to calculate:
1. Velocity of the particle.
2. Acceleration of the particle.
3. Energy associated with the motion.
4. Maximum velocity of the small bob.
#### Step-by-Step Solution:
1. Angular Frequency (ω):
[tex]\[ \omega = \frac{2\pi}{T} \][/tex]
For the particle:
[tex]\[ \omega = \frac{2\pi}{21} \approx 0.2992 \, \text{rad/s} \][/tex]
2. Velocity of the Particle (v):
[tex]\[ v = \omega \cdot A \][/tex]
Plugging in the values:
[tex]\[ v = 0.2992 \cdot 0.04 \approx 1.1968 \, \text{m/s} \][/tex]
3. Acceleration of the Particle (a):
[tex]\[ a = \omega^2 \cdot A \][/tex]
Using [tex]\(\omega \approx 0.2992 \, \text{rad/s}\)[/tex]:
[tex]\[ \omega^2 \approx 0.2992^2 \approx 0.0895 \, \text{rad}^2/\text{s}^2 \][/tex]
Therefore:
[tex]\[ a = 0.0895 \cdot 0.04 \approx 35.808 \, \text{m/s}^2 \][/tex]
4. Energy Associated with the Motion (E):
[tex]\[ E = \frac{1}{2} m \omega^2 A^2 \][/tex]
Using:
[tex]\[ \omega^2 \approx 0.0895 \, \text{rad}^2/\text{s}^2 \][/tex]
Plug in the values:
[tex]\[ E = \frac{1}{2} \cdot 0.30 \cdot 0.0895 \cdot 0.04^2 \approx 0.2148 \, \text{J} \][/tex]
5. Maximum Velocity of the Small Bob (v_max):
[tex]\[ v_{\text{max}} = A_{\text{bob}} \omega_{\text{bob}} \][/tex]
Where:
[tex]\[ \omega_{\text{bob}} = \frac{2\pi}{T_{\text{bob}}} \][/tex]
Given the same period for the bob as the particle:
[tex]\[ \omega_{\text{bob}} \approx 0.2992 \, \text{rad/s} \][/tex]
And:
[tex]\[ A_{\text{bob}} = 0.02 \, \text{m} \][/tex]
Thus:
[tex]\[ v_{\text{max}} = 0.02 \cdot 0.2992 \approx 0.0060 \, \text{m/s} \][/tex]
### Final Answers
1. Velocity of the particle: 1.1968 m/s
2. Acceleration of the particle: 35.808 m/s²
3. Energy associated with the motion: 0.2148 J
4. Maximum velocity of the small bob: 0.0060 m/s
### Definitions
#### Amplitude:
Amplitude is the maximum displacement of a particle from its equilibrium position in simple harmonic motion (SHM).
#### Period:
The period is the time taken for one complete cycle of motion.
#### Simple Pendulum:
A simple pendulum consists of a mass (often called the bob) attached to a string or rod of fixed length, swinging back and forth under the influence of gravity.
### Examples of Simple Harmonic Motion (SHM):
1. A mass-spring system, where a mass is attached to a spring and can oscillate back and forth.
2. A simple pendulum, where a mass is attached to a string and swings back and forth.
### Problem Solution:
Let's find the various quantities step-by-step:
#### Given Data:
1. Mass of the particle (m): 0.30 kg
2. Amplitude (A): 0.04 m
3. Period (T): 21 s
4. Small bob mass: 0.20 kg (or [tex]\(20 \times 10^{-2}\)[/tex] kg)
5. Small bob length: 0.02 m (or [tex]\(2 \times 10^{-2}\)[/tex] m)
6. Small bob period: 21 s
7. Small bob amplitude: 0.02 m (or [tex]\(2 \times 10^{-2}\)[/tex] m)
We need to calculate:
1. Velocity of the particle.
2. Acceleration of the particle.
3. Energy associated with the motion.
4. Maximum velocity of the small bob.
#### Step-by-Step Solution:
1. Angular Frequency (ω):
[tex]\[ \omega = \frac{2\pi}{T} \][/tex]
For the particle:
[tex]\[ \omega = \frac{2\pi}{21} \approx 0.2992 \, \text{rad/s} \][/tex]
2. Velocity of the Particle (v):
[tex]\[ v = \omega \cdot A \][/tex]
Plugging in the values:
[tex]\[ v = 0.2992 \cdot 0.04 \approx 1.1968 \, \text{m/s} \][/tex]
3. Acceleration of the Particle (a):
[tex]\[ a = \omega^2 \cdot A \][/tex]
Using [tex]\(\omega \approx 0.2992 \, \text{rad/s}\)[/tex]:
[tex]\[ \omega^2 \approx 0.2992^2 \approx 0.0895 \, \text{rad}^2/\text{s}^2 \][/tex]
Therefore:
[tex]\[ a = 0.0895 \cdot 0.04 \approx 35.808 \, \text{m/s}^2 \][/tex]
4. Energy Associated with the Motion (E):
[tex]\[ E = \frac{1}{2} m \omega^2 A^2 \][/tex]
Using:
[tex]\[ \omega^2 \approx 0.0895 \, \text{rad}^2/\text{s}^2 \][/tex]
Plug in the values:
[tex]\[ E = \frac{1}{2} \cdot 0.30 \cdot 0.0895 \cdot 0.04^2 \approx 0.2148 \, \text{J} \][/tex]
5. Maximum Velocity of the Small Bob (v_max):
[tex]\[ v_{\text{max}} = A_{\text{bob}} \omega_{\text{bob}} \][/tex]
Where:
[tex]\[ \omega_{\text{bob}} = \frac{2\pi}{T_{\text{bob}}} \][/tex]
Given the same period for the bob as the particle:
[tex]\[ \omega_{\text{bob}} \approx 0.2992 \, \text{rad/s} \][/tex]
And:
[tex]\[ A_{\text{bob}} = 0.02 \, \text{m} \][/tex]
Thus:
[tex]\[ v_{\text{max}} = 0.02 \cdot 0.2992 \approx 0.0060 \, \text{m/s} \][/tex]
### Final Answers
1. Velocity of the particle: 1.1968 m/s
2. Acceleration of the particle: 35.808 m/s²
3. Energy associated with the motion: 0.2148 J
4. Maximum velocity of the small bob: 0.0060 m/s
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