Question

Two cubes with masses `m_(1)` and `m_(2)` were interconnected by a weightless spring of stiffness `x` and placed on a smooth horizontal surface. Then the cubes were drawn closer to each other and released simultaneously. Find the natural oscillation frequency of the system.

Answer 1

In the `C.M. ` frame `(` which is rigidly attached with the centre of mass of the two cubres `)` the cubes oscillates. We know that the kinetic energy of two body system equals `(1)/(2) mu v_(rel)^(2)`, where `mu` is the reduced mass and `v_(rel)` is the modulus of velocity of any one body particle relative to other . From the conservation of mechanical energy of the oscialltion `:`
`(1)/(2) kx^(2)+ (1)/(2) mu{ (d)/(dt)(l_(0)+x)}^(2)=` constant
Here `l_(0)` is the natural length of the spring.
Differenting the above equation w.r.t. time, we get `:`
`(1)/(2) k 2 x dot(x)+(1)/(2) mu2 dot (x) ddot (x)=0[` becomes`(d(l_(0)+x))/(dt)=dot(x)]`
Thus `ddot(x)=-(k)/(mu)x(`where` mu=(m_(1)m_(2))/(m_(1)+m_(2)))`
Hence the natural frequency of oscillation `: omega_(0)=sqrt((k)/(mu))`where` mu=(m_(1)m_(2))/(m_(1)+m_(2))`