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

Step-by-step 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{(k)/(m_(equivalent))}`

`m_(equivalent)=(m_1m_2)/(m_1+m_2)`

here
`m_1=m_2=m`

`m_(eq)=(m* m)/(m+m)`

`m_(eq)=(m^2)/(2m)`

`m_(eq)=(m)/(2)`

Thus
`\omega _n=\sqrt{(k)/((m)/(2))}`

`\omega _n=\sqrt{(2k)/(m)}`