The effective mass of the spring in spring-mass system is m/3.

 The effective mass of the spring in spring-mass system is m/3.

1. Assume the Spring's Total Mass m and Length L:

Let the spring be uniformly distributed with mass m and length L. The linear mass density λ of the spring is:

$ \lambda =\frac{m}{L}$

2. Position and Velocity of Elements:

Each point on the spring moves with a different velocity during oscillations. If we assume simple harmonic motion, the velocity at a point x along the spring can be written as proportional to x.

So, if v is the velocity at the end of the spring, then the velocity of an element at position x along the spring is:

$v(x)=\frac{x}{L}.v$

3. Infinitesimal Mass Element dm:

For an infinitesimal element of length dx at position x, the mass dm is:

$dm=\lambda dx=\frac{m}{L}dx$

4. Infinitesimal Kinetic Energy:

The kinetic energy of this infinitesimal element is:

$dK=\frac{1}{2}dmv(x)^{2}=\frac{1}{2}\frac{m}{L}dx(\frac{x}{L}v)^{2}$

$dK=\frac{1}{2}\frac{m}{L}\frac{x^{2}}{L^{2}}v^{2}dx=\frac{m}{2}\frac{v^{2}}{L^{3}}x^{2}dx$

5. Total Kinetic Energy of the Spring:

Integrate dK from x = 0 to x = L to find the total kinetic energy K:

$K=\int_{0}^{L}\frac{mv^{2}}{2L^{3}}x^{2}dx$

$K=\frac{mv^{2}}{2L^{3}}\frac{L^{3}}{3}=\frac{mv^{2}}{6}$

6. Define Effective Mass:

The kinetic energy of the spring can be equated to the kinetic energy of a point mass m_eff moving with velocity v:

$\frac{1}{2}m_{eff}v^{2}=\frac{mv^{2}}{6}$

$m_{eff}=\frac{m}{3}$ 

Thus, the effective mass of the spring is indeed $\frac{m}{3}$.