A heavy table of mass M is vibrationally isolated by being hung from the ceiling by springs, so that its frequency of vertical oscillation is \omega_0 (take \omega_0 to be 2\pi/sec, a typical

value). Assume now that the ceiling vibrates vertically with amplitude A at frequency \omega\neq\omega_0, i.e. y_c(t)=A\cos(\omega t).

a. Write down the dynamical equation that relates a(t), the acceleration of the table, to its position y(t), and the position of the ceiling. Although M and k will appear in this
equation, you should be able to replace them with \omega_0. Show that the equation you get this way is the same as if a force proportional to y_c(t)=A\cos(\omega t) were acting on the mass – spring system.

This system is referred to as a driven harmonic oscillator. Its steady state solution is y_c(t)=A\cos(\omega t). NOTE that it responds solely at the drive frequency \omega\neq\omega_0, not at the
natural frequency of the oscillator \omega\neq\omega_0. (Actually there is also a transient at \omega\neq\omega_0 that fades away with time in a real system due to damping).

b. By substituting the above expression for y(t) (and the a(t) that results from this) in your equation from part (a), you should be able to obtain and solve a simple equation for C(\omega).

c. With what amplitude, y_t, will the table oscillate if the building (i.e. ceiling) oscillates with amplitude 0.01 cm at a (typical) frequency of 15 Hz? This ratio is called the isolation factor at \omega\neq\omega_0.

a. To find the dynamical equation relating the acceleration of the table, a(t), to its position, y(t), and the position of the ceiling, we can start by considering the forces acting on the table.

The restoring force from the springs can be written as F_spring = -k * y(t), where k is the spring constant. The negative sign indicates that the force is in the opposite direction of the displacement.

The force exerted by the ceiling can be written as F_ceiling = -m * a_c(t), where m is the mass of the table and a_c(t) is the acceleration of the ceiling.

By Newton's second law, the sum of these forces equals the mass of the table times its acceleration:

m * a(t) = F_spring + F_ceiling
m * a(t) = -k * y(t) - m * a_c(t)

Since we want to replace mass, m, and spring constant, k, with the angular frequency, ω_0, we can make the substitution m = ω_0^2 and k = ω_0^2.

So the equation becomes:
ω_0^2 * a(t) = -ω_0^2 * y(t) - ω_0^2 * a_c(t)

We can divide each term by ω_0^2 to simplify the equation:
a(t) = -y(t) - a_c(t)

This equation shows that the acceleration of the table, a(t), is proportional to the negative sum of the positions of the table and the ceiling.

b. To solve for the constant C(ω), we substitute the expression for y(t) into the equation from part (a) and solve for C(ω):

a(t) = -A * cos(ωt) - a_c(t)
a(t) = -A * cos(ωt) - A * cos(ωt)

Since a_c(t) is given by a_c(t) = A * cos(ωt), we can substitute it into the equation:

a(t) = -A * cos(ωt) - A * cos(ωt)
a(t) = -2A * cos(ωt)

Now, we can see that the relationship between a(t) and y(t) is linear, with a constant factor of -2A. Therefore, we can let C(ω) = -2A.

c. To find the amplitude, y_t, at which the table will oscillate when the ceiling oscillates with amplitude 0.01 cm at a frequency of 15 Hz, we can use the isolation factor.

The isolation factor, I(ω), is given by:
I(ω) = |C(ω)| / A

Here, we know that C(ω) = -2A from part (b). So the isolation factor simplifies to:
I(ω) = |-2A| / A
I(ω) = 2

Since the amplitude of the ceiling oscillation is 0.01 cm, we can multiply this by the isolation factor to find the amplitude of the table oscillation:

y_t = I(ω) * A
y_t = 2 * 0.01 cm
y_t = 0.02 cm

Therefore, the table will oscillate with an amplitude of 0.02 cm when the ceiling oscillates with an amplitude of 0.01 cm at a frequency of 15 Hz.