If mass M oscillates on a spring having mass m and stiffness k, then the natural frequency of the system is

Explanation:

Let,

S = Spring stiffness, ms = Mass of spring,m = mass of the system,V = velocity given to mass

Consider an element of length dy and mass dm at distance 'y' from the fixed point. As the variation of velocity is linear along the length of spring the velocity of an element ẏ can be written as,

\(\dot y=\text{ }\!\!~\!\!\text{ }\left( \frac{\text{V}}{\text{L}} \right)\text{y}\)

\(Mass~of~element\left( dm \right)=~\left( \frac{Mass~of~spring}{Length~of~spring} \right)\times Length~of~element\)

\(dm=\left( \frac{{{m}_{s}}}{L} \right)~dy\)

\(Kinetic~enegy~of~element=~\frac{1}{2}~dm~{{(\dot{y})}^{2}}\)

\(d{{E}_{S}}~=~\frac{1}{2}~\left( \frac{{{m}_{s}}}{L}dy \right)~{{\left( \frac{V}{L}y \right)}^{2}}\)

\(d{{E}_{S}}=~\frac{1}{2}~\left( \frac{{{m}_{s}}}{L} \right)~{{\left( \frac{V}{L}y \right)}^{2}}dy\)

\(Total~kinetic~energy~of~spring=~\int \frac{1}{2}~\left( \frac{{{m}_{s}}}{L} \right)~{{\left( \frac{V}{L}y \right)}^{2}}dy\)

\(Kinetic~energy~of~spring=~\frac{1}{2}~~\left( \frac{{{m}_{s}}}{3} \right){{V}^{2}}\)

Now the natural frequency of vibration

\(E=\frac{1}{2}~S{{x}^{2}}+\frac{1}{2}m{{V}^{2}}+\frac{1}{2}~~\left( \frac{{{m}_{s}}}{3} \right){{V}^{2}}\)

\(E=\frac{1}{2}~S{{x}^{2}}+\frac{1}{2}\left( m+\frac{{{m}_{s}}}{3}~ \right){{V}^{2}}~\)

\(\frac{{dE}}{{dt}} = \frac{1}{2}S \times 2x \times \frac{{dx}}{{dt}} + \frac{1}{2}\left( {m + \frac{{{m_s}}}{3}} \right) \times 2V \times \frac{{dV}}{{dt}}\)

\(\left( {m + \frac{{{m_s}}}{3}} \right)\ddot x + Sx = 0\)

\(~{{\omega }_{n}}=\sqrt{\frac{S}{m+\frac{{{m}_{s}}}{3}}}\)