Two blocks of mass `m_(1)` and `m_(2)` are connected with a spring of natural length `l` and spring constant k. The system is lying on a smooth horizontal surface. Initially spring is compressed by `x_(0)` as shown in figure. Show that the two blocks will perform SHM about their equilibrium position. Also (a) find the time period, (b) find amplitude of each block and (c) length of spring as a function of time.

Explanation

(a) Here both the blocks will be in equilibrium at the same time when spring is in its natural length. Let `EP_(1)` and `EP_(2)` be equilibrium positions of block A and B as shown in figure.
Let at any time during oscillations, blocks are at a distance of `x_(1)` and `x_(2)` from their equilibrium positions.
As no external force is acting on the spring block system
`:. ( m_(1) + m_(2)) Deltax_("cm") = m_(1)x_(1)-m_(2)n_(2)=0`
or `m_(1)x_(1)= m_(2)x_(2)`
For 1st particle, force equation can be written as
`k (x_(1)+(m_(1))/(m_(2))x_(1)) = -m_(1)a_(1)`
or `" " a_(1)=-(k(m_(1)+m_(2)))/(m_(1)m_(2))x_(1)`
`:. " "omega_(1)=-(k(m_(1)+m_(2)))/(m_(1)m_(2))`
Hence T = `2pi sqrt((m_(1)m_(2))/(k(m_(1)+m_(2))))=2pisqrt((mu)/(K))`
were `mu = (m_(1)m_(2))/((m_(1)+m_(2))` which is know as reduced mass
(b) Let the amplitude of blocks be `A_(1)` and `A_(2)`
`m_(1)A_(1)= m_(2)A_(2)`
By energy conservation ,
`(1)/(2)k (A_(1)+A_(2))^(2) = (1)/(2) kx_(0)^(2)`
or `" " A_(1)+A_(2) = x_(0)`
or`" " A_(1) + A_(2)= x_(0)`
or , `" " A_(1) + (m_(1))/(m_(2))A_(1)= x_(0)`
or `" " A_(1)= (m_(2)x_(0))/(m_(1)+m_(2))`
`A_(2)= (m_(1)x_(0))/(m_(1)+m_(2))`
(c) Let equilibrium position of 1st particle be origin, i.e.,
x = 0.
x co-ordinate of particles can be written as
`x_(1)= A_(1) cos omegat ` and `x_(2) = l -A_(2) cos omegat`
Hence, length of spring can be written as :
length `= x_(2)-x_(1)`
=`l-(A_(1)+A_(2)) cos omega t`