Answer :
### Examples of Systems Exhibiting Simple Harmonic Motion:
Before tackling the calculations, let’s briefly discuss two examples of systems that commonly exhibit simple harmonic motion (SHM):
1. Mass-Spring System:
Imagine a mass attached to a spring, oscillating back and forth about its equilibrium position. If the spring obeys Hooke's Law (i.e., the restoring force is proportional to the displacement), the motion of the mass is simple harmonic.
2. Pendulum:
A simple pendulum, consisting of a mass (called a bob) attached to a string of fixed length, when displaced and released, will swing back and forth in a periodic fashion. For small angular displacements, the motion approximates simple harmonic motion.
### Given Problem:
Now, let's move on to the given problem:
1. Mass of the body (m): 0.30 kg
2. Angular frequency (ω): 2.50 rad/s
3. Amplitude (A): 0.04 m
We need to calculate:
1. The maximum velocity of the particle.
2. The maximum acceleration of the particle.
3. The energy associated with the motion.
#### Step 1: Calculate the Maximum Velocity
The maximum velocity (v_max) in simple harmonic motion is given by the formula:
[tex]\[ v_{\text{max}} = \omega \cdot A \][/tex]
Given:
[tex]\[ \omega = 2.50 \, \text{rad/s} \][/tex]
[tex]\[ A = 0.04 \, \text{m} \][/tex]
Substitute the values:
[tex]\[ v_{\text{max}} = 2.50 \, \text{rad/s} \times 0.04 \, \text{m} = 0.10 \, \text{m/s} \][/tex]
So, the maximum velocity of the particle [tex]\( v_{\text{max}} \)[/tex] is 0.10 m/s.
#### Step 2: Calculate the Maximum Acceleration
The maximum acceleration (a_max) in simple harmonic motion is given by the formula:
[tex]\[ a_{\text{max}} = \omega^2 \cdot A \][/tex]
Given:
[tex]\[ \omega = 2.50 \, \text{rad/s} \][/tex]
Substitute the values:
[tex]\[ a_{\text{max}} = (2.50 \, \text{rad/s})^2 \times 0.04 \, \text{m} \][/tex]
Calculate the square of angular frequency:
[tex]\[ (2.50 \, \text{rad/s})^2 = 6.25 \, \text{rad}^2/\text{s}^2 \][/tex]
Now substitute:
[tex]\[ a_{\text{max}} = 6.25 \, \text{rad}^2/\text{s}^2 \times 0.04 \, \text{m} = 0.25 \, \text{m/s}^2 \][/tex]
So, the maximum acceleration of the particle [tex]\( a_{\text{max}} \)[/tex] is 0.25 m/s².
#### Step 3: Calculate the Energy Associated with the Motion
The total mechanical energy (E) in simple harmonic motion is given by the formula:
[tex]\[ E = \frac{1}{2} m \omega^2 A^2 \][/tex]
Given:
[tex]\[ m = 0.30 \, \text{kg} \][/tex]
[tex]\[ \omega = 2.50 \, \text{rad/s} \][/tex]
[tex]\[ A = 0.04 \, \text{m} \][/tex]
Substitute the values:
[tex]\[ E = \frac{1}{2} \times 0.30 \, \text{kg} \times (2.50 \, \text{rad/s})^2 \times (0.04 \, \text{m})^2 \][/tex]
First, calculate the square of ω:
[tex]\[ (2.50 \, \text{rad/s})^2 = 6.25 \, \text{rad}^2/\text{s}^2 \][/tex]
Next, calculate the square of A:
[tex]\[ (0.04 \, \text{m})^2 = 0.0016 \, \text{m}^2 \][/tex]
Now substitute:
[tex]\[ E = \frac{1}{2} \times 0.30 \, \text{kg} \times 6.25 \, \text{rad}^2/\text{s}^2 \times 0.0016 \, \text{m}^2 \][/tex]
Multiply the values:
[tex]\[ E = 0.5 \times 0.30 \times 6.25 \times 0.0016 \][/tex]
Simplify the multiplication:
[tex]\[ E = 0.0015 \, \text{J} \][/tex]
So, the energy associated with the motion [tex]\( E \)[/tex] is 0.0015 J.
### Summary:
- The maximum velocity of the particle is 0.10 m/s.
- The maximum acceleration of the particle is 0.25 m/s².
- The energy associated with the motion is 0.0015 J.
Before tackling the calculations, let’s briefly discuss two examples of systems that commonly exhibit simple harmonic motion (SHM):
1. Mass-Spring System:
Imagine a mass attached to a spring, oscillating back and forth about its equilibrium position. If the spring obeys Hooke's Law (i.e., the restoring force is proportional to the displacement), the motion of the mass is simple harmonic.
2. Pendulum:
A simple pendulum, consisting of a mass (called a bob) attached to a string of fixed length, when displaced and released, will swing back and forth in a periodic fashion. For small angular displacements, the motion approximates simple harmonic motion.
### Given Problem:
Now, let's move on to the given problem:
1. Mass of the body (m): 0.30 kg
2. Angular frequency (ω): 2.50 rad/s
3. Amplitude (A): 0.04 m
We need to calculate:
1. The maximum velocity of the particle.
2. The maximum acceleration of the particle.
3. The energy associated with the motion.
#### Step 1: Calculate the Maximum Velocity
The maximum velocity (v_max) in simple harmonic motion is given by the formula:
[tex]\[ v_{\text{max}} = \omega \cdot A \][/tex]
Given:
[tex]\[ \omega = 2.50 \, \text{rad/s} \][/tex]
[tex]\[ A = 0.04 \, \text{m} \][/tex]
Substitute the values:
[tex]\[ v_{\text{max}} = 2.50 \, \text{rad/s} \times 0.04 \, \text{m} = 0.10 \, \text{m/s} \][/tex]
So, the maximum velocity of the particle [tex]\( v_{\text{max}} \)[/tex] is 0.10 m/s.
#### Step 2: Calculate the Maximum Acceleration
The maximum acceleration (a_max) in simple harmonic motion is given by the formula:
[tex]\[ a_{\text{max}} = \omega^2 \cdot A \][/tex]
Given:
[tex]\[ \omega = 2.50 \, \text{rad/s} \][/tex]
Substitute the values:
[tex]\[ a_{\text{max}} = (2.50 \, \text{rad/s})^2 \times 0.04 \, \text{m} \][/tex]
Calculate the square of angular frequency:
[tex]\[ (2.50 \, \text{rad/s})^2 = 6.25 \, \text{rad}^2/\text{s}^2 \][/tex]
Now substitute:
[tex]\[ a_{\text{max}} = 6.25 \, \text{rad}^2/\text{s}^2 \times 0.04 \, \text{m} = 0.25 \, \text{m/s}^2 \][/tex]
So, the maximum acceleration of the particle [tex]\( a_{\text{max}} \)[/tex] is 0.25 m/s².
#### Step 3: Calculate the Energy Associated with the Motion
The total mechanical energy (E) in simple harmonic motion is given by the formula:
[tex]\[ E = \frac{1}{2} m \omega^2 A^2 \][/tex]
Given:
[tex]\[ m = 0.30 \, \text{kg} \][/tex]
[tex]\[ \omega = 2.50 \, \text{rad/s} \][/tex]
[tex]\[ A = 0.04 \, \text{m} \][/tex]
Substitute the values:
[tex]\[ E = \frac{1}{2} \times 0.30 \, \text{kg} \times (2.50 \, \text{rad/s})^2 \times (0.04 \, \text{m})^2 \][/tex]
First, calculate the square of ω:
[tex]\[ (2.50 \, \text{rad/s})^2 = 6.25 \, \text{rad}^2/\text{s}^2 \][/tex]
Next, calculate the square of A:
[tex]\[ (0.04 \, \text{m})^2 = 0.0016 \, \text{m}^2 \][/tex]
Now substitute:
[tex]\[ E = \frac{1}{2} \times 0.30 \, \text{kg} \times 6.25 \, \text{rad}^2/\text{s}^2 \times 0.0016 \, \text{m}^2 \][/tex]
Multiply the values:
[tex]\[ E = 0.5 \times 0.30 \times 6.25 \times 0.0016 \][/tex]
Simplify the multiplication:
[tex]\[ E = 0.0015 \, \text{J} \][/tex]
So, the energy associated with the motion [tex]\( E \)[/tex] is 0.0015 J.
### Summary:
- The maximum velocity of the particle is 0.10 m/s.
- The maximum acceleration of the particle is 0.25 m/s².
- The energy associated with the motion is 0.0015 J.
