Two balls with masses` m_(1)=1.0 kg` and `m_(2)=2.0 kg` are slipped on a thin smooth horizontal rod `(` figure`)`. The balls are interconnected by a light spring of stiffness `x=24 N//m`. The left- hand ball is imparted the initial velocity `v_(1)=12 cm//s`. Find `:`
`(a)` the oscillation frequency of the system in the process of motion,
`(b)` the energy and the amplitude of oscillations

Explanation

Suppose the balls 1 `&` 2 are displaced by `x_(1), x_(2)` from their initial position. Then the energy is `: E=(1)/(2) m_(1)x_(1)^(2)+m_(2)dot(x^(2))+m_(2)dot(x^(2))+(1)/(2)k(x_(1)-x_(2))^(2)=(1)/(2)m_(1)v_(1)^(2)`
Also total mometum is `: m_(1)dot(x_(1))+m_(2)dot(x_(2))=m_(1)v_(1)`
Define` X=(m_(1)x_(1)+m_(1)x_(2))/(m_(1)+m_(2)), x=x_(1)-x_(2)`
Then `x_(1)=X+(m_(2))/(m_(1)+m_(2))x, x_(2)=X-(m_(1))/(m_(1)+m_(2))x`
`E=(1)/(2)(m_(1)+m_(2))X^(2)+(1)/(2)(m_(1)m_(2))/(m_(1)+m_(2))dot(x^(2))+(1)/(2)kx^(2)`
Hence `dot(X)=(m_(1)v_(1))/(m_(1)+m_(2))`
So `(1)/(2)(m_(1)m_(2))/(m_(1)+m_(2))dot(x^(2))+(1)/(2) kx^(2)+(1)/(2)m_(1)v_(1)-(1)/(2)(m_(1)^(2)v_(1)^(2))/(m_(1)+m_(2))=(1)/(2)(m_(1)m_(2))/(m_(1)+m_(2))v_(1)^(2)`
`(a)` From the above equation
We see `omega=sqrt((k)/(mu))=sqrt((3xx24)/(2))=6s^(-1), ` when` mu=(m_(1)m_(2))/(m_(1)+m_(2))=(2)/(3)kg. `
`(b)` The energy of oscillation is
`(1)/(2) (m_(1)m_(2))/(m_(2)+m_(2))v_(1)^(2)=(1)/(2)(2)/(3)xx(-0.12)^(2)=48xx10^(-4)=4.8mJ`
We have `x=a sin ( omegat +alpha)`
Initially `x=0 ` at ` t=0` so `alpha=0`
Then `x=a sin omega t. ` Alsto` x=v_(1)` at `t=0`
So, `omega a =v_(1)` and hence `a=(v_(1))/(omega)=(12)/(6)=2cm`.