Question

Two identical objects of mass m are placed at either end of a spring of spring constant k and the whole system is placed on a horizontal frictionless surface.
At what angular frequency ω does the system oscillate?

Answer 1

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

$\omega _n=\sqrt{\frac{k}{m_{equivalent}}}$

$m_{equivalent}=\frac{m_1m_2}{m_1+m_2}$

here $m_1=m_2=m$

$m_{eq}=\frac{m\times m}{m+m}$

$m_{eq}=\frac{m^2}{2m}$

$m_{eq}=\frac{m}{2}$

Thus $\omega _n=\sqrt{\frac{k}{\frac{m}{2}}}$

$\omega _n=\sqrt{\frac{2k}{m}}$