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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

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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