In the final section of the lab, the student is asked to investigate the energy distribution of the spring system described previously. The student pulls the mass down an additional 16.1 cm from the equilibrium point of 21.5 cm when the mass is stationary and allows the system to oscillate. Using the equilibrium point of 21.5 cm as the zero point for total potential energy, calculate the velocity and total potential energy for each displacement given and insert the correct answer. Displacement (cm) from equilibrium Velocity (m/s) Total potential energy (J) 16.1 11.8 1.09

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Chapter1: Units, Trigonometry. And Vectors
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I have already done the first 2 parts of the question. I just want the solution of the last part.

**Title**: Understanding Simple Harmonic Motion (SHM) with a Spring

**Introduction**: 
A physics lab is demonstrating the principles of simple harmonic motion (SHM) by using a spring affixed to a horizontal support. This educational activity is designed to help students understand the concepts of spring constant and period of oscillation in SHM.

**Problem Statement**:
The student is asked to determine the spring constant (\( k \)). By suspending a mass of 295.0 g from the spring, it is observed that the spring is displaced 21.5 cm from its previous equilibrium position. Using this data, calculate the spring constant.

**Solution**:
- **Spring Constant Calculation**:
  - Spring constant (\( k \)): 13.44 N/m

**Further Analysis**:
When the spring, with the attached 295.0 g mass, is displaced from its new equilibrium position, it undergoes simple harmonic motion. Calculate the period of oscillation (\( T \)), neglecting the mass of the spring itself.

- **Period of Oscillation**:
  - \( T = 0.93 \) seconds

**Conclusion**:
This exercise illustrates the basic principles of SHM, providing practical insight into the calculation of spring constants and oscillation periods using experimental data. Through this example, students gain a deeper understanding of the mechanics behind SHM, essential in various physics applications.
Transcribed Image Text:**Title**: Understanding Simple Harmonic Motion (SHM) with a Spring **Introduction**: A physics lab is demonstrating the principles of simple harmonic motion (SHM) by using a spring affixed to a horizontal support. This educational activity is designed to help students understand the concepts of spring constant and period of oscillation in SHM. **Problem Statement**: The student is asked to determine the spring constant (\( k \)). By suspending a mass of 295.0 g from the spring, it is observed that the spring is displaced 21.5 cm from its previous equilibrium position. Using this data, calculate the spring constant. **Solution**: - **Spring Constant Calculation**: - Spring constant (\( k \)): 13.44 N/m **Further Analysis**: When the spring, with the attached 295.0 g mass, is displaced from its new equilibrium position, it undergoes simple harmonic motion. Calculate the period of oscillation (\( T \)), neglecting the mass of the spring itself. - **Period of Oscillation**: - \( T = 0.93 \) seconds **Conclusion**: This exercise illustrates the basic principles of SHM, providing practical insight into the calculation of spring constants and oscillation periods using experimental data. Through this example, students gain a deeper understanding of the mechanics behind SHM, essential in various physics applications.
In the final section of the lab, the student is asked to investigate the energy distribution of the spring system described previously. The student pulls the mass down an additional 16.1 cm from the equilibrium point of 21.5 cm when the mass is stationary and allows the system to oscillate. Using the equilibrium point of 21.5 cm as the zero point for total potential energy, calculate the velocity and total potential energy for each displacement given and insert the correct answer.

The table displayed includes the following columns:

- **Displacement (cm) from equilibrium**
- **Velocity (m/s)**
- **Total potential energy (J)**

| Displacement (cm) from equilibrium | Velocity (m/s) | Total potential energy (J) |
|-----------------------------------|----------------|-----------------------------|
| 16.1                              | 0              |                             |
| 11.8                              | 1.09           |                             |

There is an "Answer Bank" provided with the following values to choose from:

- 0.311
- 0.0941
- 0.175
- 0
- 0.74

Students must calculate and fill in the total potential energy values for the given displacements.
Transcribed Image Text:In the final section of the lab, the student is asked to investigate the energy distribution of the spring system described previously. The student pulls the mass down an additional 16.1 cm from the equilibrium point of 21.5 cm when the mass is stationary and allows the system to oscillate. Using the equilibrium point of 21.5 cm as the zero point for total potential energy, calculate the velocity and total potential energy for each displacement given and insert the correct answer. The table displayed includes the following columns: - **Displacement (cm) from equilibrium** - **Velocity (m/s)** - **Total potential energy (J)** | Displacement (cm) from equilibrium | Velocity (m/s) | Total potential energy (J) | |-----------------------------------|----------------|-----------------------------| | 16.1 | 0 | | | 11.8 | 1.09 | | There is an "Answer Bank" provided with the following values to choose from: - 0.311 - 0.0941 - 0.175 - 0 - 0.74 Students must calculate and fill in the total potential energy values for the given displacements.
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