Calculate The Period Of Vibration Of A Spring System With A Force Constant Of $200 \, \text{Nm}^{-1}$ When Loaded With A Mass Of $2 \, \text{kg}$.A. 0.53 S B. 0.63 S C. 0.33 S D.

Introduction

In physics, the study of vibrations and oscillations is crucial in understanding various phenomena, including the behavior of springs and masses. The period of vibration of a spring system is a fundamental concept that can be calculated using the force constant of the spring and the mass attached to it. In this article, we will explore how to calculate the period of vibration of a spring system with a force constant of $200 \, \text{Nm}^{-1}$ when loaded with a mass of $2 \, \text{kg}$.

Understanding the Basics

The period of vibration of a spring system is the time taken by the system to complete one oscillation or cycle. It is a measure of the time it takes for the system to return to its original position after being displaced. The period of vibration is denoted by the symbol $T$ and is measured in seconds.

The force constant of a spring, denoted by the symbol $k$, is a measure of the stiffness of the spring. It is defined as the force required to displace the spring by a unit distance. The force constant is measured in units of Newtons per meter (Nm$^{-1}$).

The mass of the object attached to the spring, denoted by the symbol $m$, is a measure of the inertia of the object. It is defined as the amount of matter in the object and is measured in units of kilograms (kg).

Derivation of the Period of Vibration Formula

The period of vibration of a spring system can be calculated using the following formula:

$$T = 2\pi \sqrt{\frac{m}{k}}$$

where $T$ is the period of vibration, $m$ is the mass of the object attached to the spring, and $k$ is the force constant of the spring.

Applying the Formula

Now that we have the formula, let's apply it to the given problem. We are given a spring system with a force constant of $200 \, \text{Nm}^{-1}$ and a mass of $2 \, \text{kg}$. We need to calculate the period of vibration of this system.

Plugging in the values, we get:

$$T = 2\pi \sqrt{\frac{2}{200}}$$

Simplifying the expression, we get:

$$T = 2\pi \sqrt{\frac{1}{100}}$$

$$T = 2\pi \times \frac{1}{10}$$

$$T = \frac{\pi}{5}$$

$$T = 0.63 \, \text{s}$$

Conclusion

In this article, we have calculated the period of vibration of a spring system with a force constant of $200 \, \text{Nm}^{-1}$ when loaded with a mass of $2 \, \text{kg}$. We have used the formula $T = 2\pi \sqrt{\frac{m}{k}}$ to derive the period of vibration and have applied it to the given problem. The calculated period of vibration is $0.63 \, \text{s}$.

Comparison with Options

Now that we have calculated the period of vibration, let's compare it with the given options:

• A. $0.53 \, \text{s}$
• B. $0.63 \, \text{s}$
• C. $0.33 \, \text{s}$

Our calculated period of vibration, $0.63 \, \text{s}$, matches option B.

Discussion

The period of vibration of a spring system is an important concept in physics that can be used to understand various phenomena, including the behavior of springs and masses. The formula $T = 2\pi \sqrt{\frac{m}{k}}$ can be used to calculate the period of vibration of a spring system given the force constant and the mass attached to it.

In this article, we have calculated the period of vibration of a spring system with a force constant of $200 \, \text{Nm}^{-1}$ when loaded with a mass of $2 \, \text{kg}$. We have used the formula to derive the period of vibration and have applied it to the given problem. The calculated period of vibration is $0.63 \, \text{s}$.

Limitations

The formula $T = 2\pi \sqrt{\frac{m}{k}}$ assumes that the spring is ideal and that there are no external forces acting on the system. In reality, the spring may not be ideal, and there may be external forces acting on the system, which can affect the period of vibration.

Future Work

In future work, we can explore the effects of non-ideal springs and external forces on the period of vibration of a spring system. We can also investigate the behavior of more complex systems, such as systems with multiple springs and masses.

References

• [1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons.
• [2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.

Appendix

The following is a list of formulas and equations used in this article:

• $T = 2\pi \sqrt{\frac{m}{k}}$
• $k = \frac{F}{x}$
• $F = kx$

Introduction

In our previous article, we calculated the period of vibration of a spring system with a force constant of $200 \, \text{Nm}^{-1}$ when loaded with a mass of $2 \, \text{kg}$. We used the formula $T = 2\pi \sqrt{\frac{m}{k}}$ to derive the period of vibration and applied it to the given problem. In this article, we will answer some frequently asked questions related to calculating the period of vibration of a spring system.

Q: What is the period of vibration of a spring system?

A: The period of vibration of a spring system is the time taken by the system to complete one oscillation or cycle. It is a measure of the time it takes for the system to return to its original position after being displaced.

Q: What is the formula for calculating the period of vibration of a spring system?

A: The formula for calculating the period of vibration of a spring system is:

$$T = 2\pi \sqrt{\frac{m}{k}}$$

where $T$ is the period of vibration, $m$ is the mass of the object attached to the spring, and $k$ is the force constant of the spring.

Q: What is the force constant of a spring?

A: The force constant of a spring is a measure of the stiffness of the spring. It is defined as the force required to displace the spring by a unit distance. The force constant is measured in units of Newtons per meter (Nm$^{-1}$).

Q: What is the mass of an object?

A: The mass of an object is a measure of the inertia of the object. It is defined as the amount of matter in the object and is measured in units of kilograms (kg).

Q: How do I calculate the period of vibration of a spring system with a force constant of $300 \, \text{Nm}^{-1}$ and a mass of $3 \, \text{kg}$?

A: To calculate the period of vibration of a spring system with a force constant of $300 \, \text{Nm}^{-1}$ and a mass of $3 \, \text{kg}$, you can use the formula:

$$T = 2\pi \sqrt{\frac{m}{k}}$$

Plugging in the values, you get:

$$T = 2\pi \sqrt{\frac{3}{300}}$$

Simplifying the expression, you get:

$$T = 2\pi \sqrt{\frac{1}{100}}$$

$$T = 2\pi \times \frac{1}{10}$$

$$T = \frac{\pi}{5}$$

$$T = 0.63 \, \text{s}$$

Q: What are some common mistakes to avoid when calculating the period of vibration of a spring system?

A: Some common mistakes to avoid when calculating the period of vibration of a spring system include:

• Using the wrong formula
• Plugging in the wrong values
• Not simplifying the expression
• Not checking the units

Q: Can I use the formula to calculate the period of vibration of a spring system with a non-ideal spring?

A: No, the formula assumes that the spring is ideal and that there are no external forces acting on the system. If the spring is non-ideal or there are external forces acting on the system, you will need to use a more complex formula or consult with a physicist.

Conclusion

In this article, we have answered some frequently asked questions related to calculating the period of vibration of a spring system. We have provided the formula for calculating the period of vibration and have answered questions related to the force constant, mass, and period of vibration. We have also discussed some common mistakes to avoid when calculating the period of vibration of a spring system.

References

• [1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons.
• [2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.

Appendix

The following is a list of formulas and equations used in this article:

• $T = 2\pi \sqrt{\frac{m}{k}}$
• $k = \frac{F}{x}$
• $F = kx$

Note: The formulas and equations used in this article are based on the principles of physics and are widely accepted in the scientific community.