Answer:
Explanation:
It is given that two identical masses is attached to the either end of the spring
with spring constant k
angular frequency of the system is given by
[tex]\omega _n=\sqrt{\frac{k}{m_{equivalent}}}[/tex]
[tex]m_{equivalent}=\frac{m_1m_2}{m_1+m_2}[/tex]
here [tex]m_1=m_2=m[/tex]
[tex]m_{eq}=\frac{m\times m}{m+m}[/tex]
[tex]m_{eq}=\frac{m^2}{2m}[/tex]
[tex]m_{eq}=\frac{m}{2}[/tex]
Thus [tex]\omega _n=\sqrt{\frac{k}{\frac{m}{2}}}[/tex]
[tex]\omega _n=\sqrt{\frac{2k}{m}}[/tex]
