The Effect of the Mass of the Spring on the Natural Frequency of the System

Statement: Determine the effect of the mass of the spring on the natural frequency of the system shown in figure - 1.  

Solution:

Let x be the displacement of mass m and so the velocity will be `\dot{x}`. The velocity of spring element at a distance y from the fixed end maybe written as `\dot{x}\frac{y}{l}`.

Where `l` is the total length of the spring.

The kinetic energy of spring element `dy` per unit area is written as, 

`(K.E)_s` = `\frac{1}{2}(\rho dy)(\dot{x}\frac{y}{l})^2` ............................. (i)

And, the kinetic energy of the mass m,

`(K.E)_m` = `\frac{1}{2}\ m\ {\dot{x}}^2` ................................... (ii)

Hence, for the interval `y = 0` to `y = l`, Total kinetic energy of the spring mass system,

T. K. E = `(K.E)_m` + `(K.E)_s`

        = ` \frac{1}{2}m{\dot{x}}^2+\frac{1}{2l^2}\rho{\dot{x}}^2 \int_{0}^{l}{y^2dy}`

        =`\frac{1}{2}m{\dot{x}}^2+\frac{1}{2l^2}\ \rho{\dot{x}}^2 [\frac{y^3}{3}\right]_0^l`

        = `\frac{1}{2}m{\dot{x}}^2+\frac{1}{2l^2}\rho{\dot{x}}^2 \frac{l^3}{3}`

        = `\frac{1}{2}m{\dot{x}}^2+\frac{1}{6}\ \rho{\dot{x}}^2 l`

        =`\frac{1}{2} (m+\frac{\rho l}{3}){\dot{x}}^2`

Therefore, 

T.K.E = `\frac{1}{2} (m+\frac{\rho l}{3}){\dot{x}}^2`  .................................... (iii) 

The potential energy of spring element `dy` is written as, 

`(P.E)_s` = `\frac{1}{2} kx^2`

The potential energy of the mass,

`(P.E)_m` = `mg \times 0=0` 

Therefore, Total Potential Energy of the system written as,

T.P.E = `(P.E)_s+(P.E)_m=\frac{1}{2}kx^2` ............................... (iv) 

Hence, Total Energy = Total Potential Energy + Total Kinetic Energy

`E_(Total)`=`\frac{1}{2}kx^2+\frac{1}{2} (m+\frac{\rho l}{3}){\dot{x}}^2` ............................. (v)

The total energy of the system is constant. Therefore,

`E_{Total}` = `Constant`

`\Rightarrow \frac{1}{2}kx^2+\frac{1}{2}(m+\frac{\rho l}{3}\right){\dot{x}}^2`=`Constant`.

`\Rightarrow \frac{d}{dt} [\frac{1}{2}kx^2+\frac{1}{2} (m+\frac{\rho l}{3} ){\dot{x}}^2 ]` = `0`

`\Rightarrow kx+ (m+\frac{\rho l}{3})\ddot{x}` = `0`

`\Rightarrow (m+\frac{\rho l}{3} )\ddot{x}+kx` = `0` ............................... (vi) 

This is the differential equation of motion of the spring mass system.

If the motion is simple harmonic, then let, the solution of this second order differential equation is,
`x=A \sin{\omega_n t}`

Where,
`\omega_n` = natural frequency
A = Maximum displacement of the mass

Therefore, 

`\Rightarrow (m+\frac{\rho l}{3} ) (\omega_n )^2A\sin{\omega_n t}+k\ A\sin{\omega_n t}` = `0`

`\Rightarrow (m+\frac{\rho l}{3} ) (\omega_n)^2+k` = `0`

`\Rightarrow (\omega_n )^2` = `\frac{k}{m+\frac{\rho l}{3}}`

`\therefore\omega_n` = `\sqrt{\frac{k}{m+\frac{\rho l}{3}}}` ............................................ (vii)

The natural frequency, `\omega_n` = `\sqrt{\frac{k}{m+\frac{\rho l}{3}}}`

Hence, we can conclude if the mass m is increased the natural frequency can be decreased.