Question

Consider N identical masses connected through springs of same force constant k. The free ends of the coupled system are rigidly fixed at x=0 and x=L The masses are made to execute longitudinal oscillations.
i) Depict the equilibrium configuration as well as instantaneous configuration.
ii) Derive the equation of wave motion.
iii) Calculate the velocity of wave when a force of 500 N acts on a system with mass per unit length 0.25 kg/m

Answer 1

Suppose the length of each spring in its natural unextended or uncompressed form is   a.  Then the total length of  N+1 springs is  (N+1) a. 

1)
In the equilibrium position, the springs are not oscillating.  Each is extended by:
      Δ  =  L /(N+1)  - a
On either side of each mass the force acting on it is:  k Δ
Mass1 is at  x =  a + Δ
mass 2 is at  x = 2 a + 2 Δ
mass N  is at  x =  N a + NΔ

2)    longitudinal oscillations

Let the displacements to the right (positive x axis direction) be x1, x2, x3 ... xn of the blocks.

Equations of motion of the masses are:
  m d²x1 /dt² = - k(2 x1 - x2)
  m d²x2 /dt²  = - k(2 x2 - x1 - x3)
  m  d² x3 / dt² = - k(2 x3 - x2 - x4)
     :      :                :
  m d² x_n-1 /dt²  = -  k (2 x_n-1  - x_n-2  - x_n )
  m  d² x_n / dt²  = - k (2 x_n  - x_n-1)


Suppose we assume that  x_n (t) = A_n  Cos ωt
                   =>  x_n" (t) = - ω²  x_n(t)

Let  2 - m ω²/k  = 2 - (ω/ω₀) = C

we get equations like:
 C x1  = x2
  C x2 = x1 + x3
  C x3 = x2 + x4
  C x4 =  x3  + x5          
   :   
  C  x_n-1 =  x_n-2  + x_n
   C   x_n =   x_n-1

the solutions are:    we assume  x_i =  A_i  Cos ωt
  C = 0.5549  or - 0.8019  for  4 mass system...  and   ω/ω₀  = 1.202  or   1.673 
          
 if there are 3 masses then:  C = √2 or  -√2 
                     ω/ω₀ =  2 + √2    or  2 - √2

3)
longitudinal wave speed along the medium
 speed is = √(F/μ )  =  √(500/0.25) = 20 √5 m/sec