Answer :
Sure! Let's tackle each part of the question step-by-step:
### Part (c)
#### i. Maximum Velocity of the Particle
To find the maximum velocity ([tex]\(v_{\text{max}}\)[/tex]) of the particle in simple harmonic motion, we use the given speed value. For simplicity, we can directly use the provided speed as the maximum velocity:
[tex]\[ v_{\text{max}} = 2.505 \, \text{m/s} \][/tex]
#### ii. Maximum Acceleration of the Particle
The maximum acceleration ([tex]\(a_{\text{max}}\)[/tex]) in simple harmonic motion is given by [tex]\(a_{\text{max}} = \omega^2 \cdot \text{amplitude}\)[/tex], where [tex]\(\omega\)[/tex] is the angular frequency.
To find [tex]\(\omega\)[/tex], we use:
[tex]\[ \omega = \frac{v_{\text{max}}}{\text{amplitude}} \][/tex]
Given:
- [tex]\(v_{\text{max}} = 2.505 \, \text{m/s}\)[/tex]
- [tex]\(\text{amplitude} = 0.04 \, \text{m}\)[/tex]
Calculate [tex]\(\omega\)[/tex]:
[tex]\[ \omega = \frac{2.505}{0.04} \approx 62.625 \, \text{rad/s} \][/tex]
Now, calculate [tex]\(a_{\text{max}}\)[/tex]:
[tex]\[ a_{\text{max}} = \omega^2 \cdot \text{amplitude} = (62.625)^2 \cdot 0.04 \approx 156.876 \, \text{m/s}^2 \][/tex]
#### iii. Energy Associated with the Motion
The total mechanical energy ([tex]\(E\)[/tex]) in simple harmonic motion is given by:
[tex]\[ E = \frac{1}{2} k \cdot \text{amplitude}^2 \][/tex]
where [tex]\(k\)[/tex] is the spring constant. Using the relationship [tex]\(v_{\text{max}} = \omega \cdot \text{amplitude}\)[/tex] and [tex]\(\omega = \sqrt{\frac{k}{m}}\)[/tex], we can find [tex]\(k\)[/tex] as follows:
[tex]\[ v_{\text{max}}^2 = k/m \cdot \text{amplitude}^2 \implies k = \frac{m \cdot v_{\text{max}}^2}{\text{amplitude}^2} \][/tex]
Given:
- [tex]\(m = 0.30 \, \text{kg}\)[/tex]
- [tex]\(v_{\text{max}} = 2.505 \, \text{m/s}\)[/tex]
- [tex]\(\text{amplitude} = 0.04 \, \text{m}\)[/tex]
Calculate [tex]\(k\)[/tex]:
[tex]\[ k = \frac{0.30 \cdot (2.505)^2}{(0.04)^2} \approx 117.656 \, \text{N/m} \][/tex]
Now, calculate [tex]\(E\)[/tex]:
[tex]\[ E = \frac{1}{2} \cdot 117.656 \cdot (0.04)^2 \approx 0.941 \, \text{J} \][/tex]
### Part (d)
#### Calculate the Maximum Velocity of the Bob
The maximum velocity ([tex]\(v_{\text{max}}\)[/tex]) of the bob in simple harmonic motion is given by:
[tex]\[ v_{\text{max}} = \omega \cdot \text{amplitude} \][/tex]
First, we find the angular frequency [tex]\(\omega\)[/tex] using the period [tex]\(T\)[/tex]:
[tex]\[ \omega = \frac{2\pi}{T} \][/tex]
Given:
- [tex]\(T = 2.1 \, \text{s}\)[/tex]
- [tex]\(\text{amplitude} = 5.0 \times 10^{-2} \, \text{m}\)[/tex]
Calculate [tex]\(\omega\)[/tex]:
[tex]\[ \omega = \frac{2\pi}{2.1} \approx 2.994 \, \text{rad/s} \][/tex]
Now, calculate [tex]\(v_{\text{max}}\)[/tex]:
[tex]\[ v_{\text{max}} = 2.994 \cdot 0.05 \approx 0.150 \, \text{m/s} \][/tex]
Thus, the final results are:
- Maximum velocity of the particle (part c): [tex]\(2.505 \, \text{m/s}\)[/tex]
- Maximum acceleration of the particle: [tex]\(156.876 \, \text{m/s}^2\)[/tex]
- Energy associated with the motion: [tex]\(0.941 \, \text{J}\)[/tex]
- Maximum velocity of the bob (part d): [tex]\(0.150 \, \text{m/s}\)[/tex]
### Part (c)
#### i. Maximum Velocity of the Particle
To find the maximum velocity ([tex]\(v_{\text{max}}\)[/tex]) of the particle in simple harmonic motion, we use the given speed value. For simplicity, we can directly use the provided speed as the maximum velocity:
[tex]\[ v_{\text{max}} = 2.505 \, \text{m/s} \][/tex]
#### ii. Maximum Acceleration of the Particle
The maximum acceleration ([tex]\(a_{\text{max}}\)[/tex]) in simple harmonic motion is given by [tex]\(a_{\text{max}} = \omega^2 \cdot \text{amplitude}\)[/tex], where [tex]\(\omega\)[/tex] is the angular frequency.
To find [tex]\(\omega\)[/tex], we use:
[tex]\[ \omega = \frac{v_{\text{max}}}{\text{amplitude}} \][/tex]
Given:
- [tex]\(v_{\text{max}} = 2.505 \, \text{m/s}\)[/tex]
- [tex]\(\text{amplitude} = 0.04 \, \text{m}\)[/tex]
Calculate [tex]\(\omega\)[/tex]:
[tex]\[ \omega = \frac{2.505}{0.04} \approx 62.625 \, \text{rad/s} \][/tex]
Now, calculate [tex]\(a_{\text{max}}\)[/tex]:
[tex]\[ a_{\text{max}} = \omega^2 \cdot \text{amplitude} = (62.625)^2 \cdot 0.04 \approx 156.876 \, \text{m/s}^2 \][/tex]
#### iii. Energy Associated with the Motion
The total mechanical energy ([tex]\(E\)[/tex]) in simple harmonic motion is given by:
[tex]\[ E = \frac{1}{2} k \cdot \text{amplitude}^2 \][/tex]
where [tex]\(k\)[/tex] is the spring constant. Using the relationship [tex]\(v_{\text{max}} = \omega \cdot \text{amplitude}\)[/tex] and [tex]\(\omega = \sqrt{\frac{k}{m}}\)[/tex], we can find [tex]\(k\)[/tex] as follows:
[tex]\[ v_{\text{max}}^2 = k/m \cdot \text{amplitude}^2 \implies k = \frac{m \cdot v_{\text{max}}^2}{\text{amplitude}^2} \][/tex]
Given:
- [tex]\(m = 0.30 \, \text{kg}\)[/tex]
- [tex]\(v_{\text{max}} = 2.505 \, \text{m/s}\)[/tex]
- [tex]\(\text{amplitude} = 0.04 \, \text{m}\)[/tex]
Calculate [tex]\(k\)[/tex]:
[tex]\[ k = \frac{0.30 \cdot (2.505)^2}{(0.04)^2} \approx 117.656 \, \text{N/m} \][/tex]
Now, calculate [tex]\(E\)[/tex]:
[tex]\[ E = \frac{1}{2} \cdot 117.656 \cdot (0.04)^2 \approx 0.941 \, \text{J} \][/tex]
### Part (d)
#### Calculate the Maximum Velocity of the Bob
The maximum velocity ([tex]\(v_{\text{max}}\)[/tex]) of the bob in simple harmonic motion is given by:
[tex]\[ v_{\text{max}} = \omega \cdot \text{amplitude} \][/tex]
First, we find the angular frequency [tex]\(\omega\)[/tex] using the period [tex]\(T\)[/tex]:
[tex]\[ \omega = \frac{2\pi}{T} \][/tex]
Given:
- [tex]\(T = 2.1 \, \text{s}\)[/tex]
- [tex]\(\text{amplitude} = 5.0 \times 10^{-2} \, \text{m}\)[/tex]
Calculate [tex]\(\omega\)[/tex]:
[tex]\[ \omega = \frac{2\pi}{2.1} \approx 2.994 \, \text{rad/s} \][/tex]
Now, calculate [tex]\(v_{\text{max}}\)[/tex]:
[tex]\[ v_{\text{max}} = 2.994 \cdot 0.05 \approx 0.150 \, \text{m/s} \][/tex]
Thus, the final results are:
- Maximum velocity of the particle (part c): [tex]\(2.505 \, \text{m/s}\)[/tex]
- Maximum acceleration of the particle: [tex]\(156.876 \, \text{m/s}^2\)[/tex]
- Energy associated with the motion: [tex]\(0.941 \, \text{J}\)[/tex]
- Maximum velocity of the bob (part d): [tex]\(0.150 \, \text{m/s}\)[/tex]
