Question

Two equal masses are connected by means of a spring and two other identical springs fixed to rigid supports on either side (Fig. 1.3), permit the masses to jointly undergo simple harmonic motion along a straight line, so that the system corresponds to two coupled oscillations. 

Assume that m₁ = m₂ = m and the stiffness constant is k for both the oscillators. 

(a) Form the differential equations for both the oscillators and solve the coupled equations and obtain the frequencies of oscillations. 

(b) Discuss the modes of oscillation and sketch the modes.

Answer 1

(a) Let the mass 1 be displaced by x₁ and mass m₂ by x₂. The force due to the spring on the left acting on mass 1 is −kx₁ and that due to the coupling is −k(x₁ − x₂). 

The net force 

F₁ = −kx₁ − k(x₁ − x₂) = −k(2x₁ − x₂) 

The equation of motion for mass 1 is mx₁ + k(2x₁ − x₂) = 0 (1) 

Similarly, for mass 2, the spring on the right exerts a force −kx₂, and the coupling spring exerts a force −k(x₂ − x₁). The net force 

F₂ = −kx₂ − k(x₂ − x₁) = −k(2x₂ − x₁)

In order that the above equations may have a non-trivial solution, the determinant formed from the coefficients of x₁ and x₂ must vanish.

In (12) and (13) the amplitudes are not all independent as we can verify with the use of (7) and (8). Substituting (10) and (12) in (7), yields A₂ = A₁. Substitution of (11) and (13) in (7), gives B₂ = −B₁. 

Dropping off the subscripts on A s and B s the solutions can be written as

If initially x₁ = x₂, the masses oscillate in phase with frequency ω₁ (symmetrical mode) as in Fig. 1.16(a). If initially x₂ = −x₂ then the masses oscillate out of phase (asymmetrical) as in Fig. 1.16(b)