# 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.