two springs and two masses are attached in a straight line on a horizontal frictionless surface. the system is set in motion by given external force f(t) acting on the right hand mass m2. we denoted by x(t)the displacement(to the right) of mass m1 from static equilibrium position, and by y(t) the displacement of the mass m1 from static its equilibrium position. the two springs are neither stretched nor compressed when x(t) and y(t) are zero.

Question

two springs and two masses are attached in a straight line on a horizontal frictionless surface. the system is set in motion by given external force f(t) acting on the right hand mass m2. we denoted by x(t)the displacement(to the right) of mass m1 from static equilibrium position, and by y(t) the displacement of the mass m1 from static its equilibrium position. the two springs are neither stretched nor compressed when x(t) and y(t) are zero.

a) find the equation of the system
b)determine the two normal frequencies for the system.

Answer 1

a) To find the equation of the system, we need to consider the forces acting on each mass and write down the equations of motion.

Let's denote the displacement of the mass m1 from static equilibrium position as x(t), and the displacement of the mass m2 from its static equilibrium position as y(t). The equations of motion for mass m1 and m2 can be written as follows:

For mass m1:
m1 * d²x/dt² = -k1 * x(t) - k2 * (x(t) - y(t))

For mass m2:
m2 * d²y/dt² = f(t) - k2 * (y(t) - x(t))

Here, m1 and m2 are the masses of m1 and m2 respectively, k1 and k2 are the spring constants of the two springs, and f(t) is the external force acting on mass m2.

b) To determine the two normal frequencies of the system, we can solve the equations of motion by assuming harmonic motion solutions.

Let's assume harmonic solutions of the form:

x(t) = X * cos(ωt + φ)
y(t) = Y * cos(ωt + φ)

Here, X and Y are the amplitudes of the motions, ω is the angular frequency, and φ is the phase angle.

Substituting these harmonic solutions into the equations of motion, we get:

For mass m1:
-m1 * ω² * X * cos(ωt + φ) = -k1 * X * cos(ωt + φ) - k2 * (X * cos(ωt + φ) - Y * cos(ωt + φ))
Simplifying, we get:
ω²(m1 + (k1 + k2)) * X = k2 * Y

For mass m2:
-m2 * ω² * Y * cos(ωt + φ) = -k2 * (Y * cos(ωt + φ) - X * cos(ωt + φ)) + f(t)
Simplifying, we get:
ω²(m2 + k2) * Y = f(t) + k2 * X

From these equations, we can see that the normal frequencies are given by the square root of the ratios ω²(m1 + (k1 + k2)) / k2 and ω²(m2 + k2) / k2.

Therefore, the two normal frequencies for the system are:
ω1 = sqrt((m1 + (k1 + k2)) / (m2 + k2))
ω2 = sqrt(m2 / (m2 + k2))