An ideal spring is connected to the ceiling as shown above. A mass M=7kg is connected to the bottom end of the spring and is allowed to reach a new equilibrium position, stretching the spring a displacement L=0.075m. The spring is then raised from the equilibrium position to a height ymax = 0.035 m and released. What is the position of the mass relative to equilibrium at t=1.1 (treat upward as the +y direction)?

To solve this problem, we need to use the formula for the position of an oscillating spring-mass system, which is given by:

y(t) = A cos(ωt + φ)

where:

• A is the amplitude of the oscillation, which is the maximum displacement from the equilibrium position (in this case, ymax = 0.035 m),
• ω is the angular frequency of the oscillation, which is related to the spring constant k and the mass M by the formula ω = sqrt(k/M),
• t is the time (in this case, t = 1.1 s), and
• φ is the phase constant, which depends on the initial conditions of the system.

First, we need to find the spring constant k. We know that the force exerted by the spring when it is stretched by a displacement L is given by Hooke's law:

F = kL

We also know that the force exerted by the spring must balance the weight of the mass M at the equilibrium position, so we have:

Mg = kL

Solving for k, we get:

k = Mg/L = (7 kg)(9.8 m/s²)/(0.075 m) = 917.33 N/m

Next, we can find the angular frequency ω:

ω = sqrt(k/M) = sqrt((917.33 N/m)/(7 kg)) = 11.43 rad/s

Since the mass is released from the maximum displacement with zero initial velocity, the phase constant φ is 0.

Finally, we can substitute these values into the formula for y(t) to find the position of the mass at t = 1.1 s:

y(1.1 s) = (0.035 m) cos((11.43 rad/s)(1.1 s) + 0) = -0.027 m

So, the position of the mass relative to the equilibrium at t = 1.1 s is -0.027 m, which means it is 0.027 m below the equilibrium position.