To find the natural frequency of a spring-mass system considering the mass of the spring, we need to understand how the spring's mass contributes to the system's dynamics.

Step 1: Understanding the System

A typical spring-mass system without considering the spring's mass is characterized by the formula for natural frequency:

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

where:

• $f$ is the natural frequency,
• $k$ is the spring constant, and
• $m$ is the mass of the attached object.

Step 2: Considering the Spring's Mass

When the spring has mass, it cannot be treated simply as a massless element. The mass of the spring itself contributes to the inertia of the system. However, because different parts of the spring move with different velocities, the total kinetic energy isn't just that of a single lumped mass but rather distributed across the length of the spring.

Step 3: Effective Mass of the Spring

Through detailed analysis, it can be shown that the effect of the spring's mass on the system is equivalent to adding an additional mass to the system. This additional mass is equal to one-third of the spring's mass ($m_s$).

So, the effective total mass $m_{\text{eff}}$ for the system becomes:

$m_{\text{eff}} = m + \frac{1}{3}m_s$

Step 4: Modified Natural Frequency

Substituting this effective mass into the natural frequency formula:

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m + \frac{1}{3}m_s}}$

Conclusion

The above equation shows that when the mass of the spring is considered, it is equivalent to adding one-third of the spring's mass to the main mass for the purpose of calculating the natural frequency. This adjustment accounts for the distributed inertia of the spring.

This result aligns with the physical intuition that a portion of the spring's mass acts as if it were concentrated at the point where the mass is attached.

---

Do you have any questions or would you like more details on a specific part of the explanation? Here are some related questions to consider:

1. How does the spring constant $k$ affect the natural frequency of the system?
2. What assumptions are made in deriving the effective mass of the spring?
3. How would the system's behavior change if the spring's mass were much larger than the attached mass?
4. Can you derive the expression for the effective mass of the spring?
5. What are the implications of this result in practical engineering applications?

Tip: When considering the mass of the spring in practical problems, always check if it is significant compared to the attached mass. If it is, the $\frac{1}{3}$ correction can lead to more accurate results.