From health tips to tech hacks, find it all on IDNLearn.com. Ask your questions and receive comprehensive and trustworthy answers from our experienced community of professionals.
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
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.
