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Chapter 2: Problem 19

A spring fixed at its upper end is stretched 6 inches by a 10-pound weight attached at its lower end. The spring-mass system is suspended in a viscous medium so that the system is subjected to a damping force of \(5 \frac{d x}{d t}\) lbs. Describe the motion of the system if the weight is drawn down an additional 4 inches and released. What would happen if you changed the coefficient " 5 " to "4"? [You may need to consult your introductory physics text.]

Short Answer

Expert verified
Given that the spring constant, damping ratio and the additional displacement have been determined, the motion of the system can be described by solving the formed ordinary differential equation. This will unveil the nature of the oscillations: whether the system is over-damped, under-damped or critically-damped. Changing the damping coefficient from '5' to '4' affects the damping ratio, which would in turn alter the system's response - potentially changing its damping status (becoming over, under or critically damped). The exact results depend on the numerical values of the parameters involved and the actual calculation made.

Step by step solution

01

Determine the spring constant

Find \(k\) by applying Hooke's Law, \(F=kx\), where \(F\) is the force applied (weight 10 pounds), and \(x\) is the displacement (6 inches or 0.5 feet). Solve to get the spring constant \(k = \frac{F}{x}\).
02

Calculate the damping ratio

The damping force is \(5 \frac{dx}{dt}\). The damping ratio, \(\zeta\), can be found by dividing the damping force by the force obtained through multiplying the mass \(m\) (which in this case is the weight divided by gravitational acceleration) and 2 times the square root of the spring constant, \( \zeta = \frac{5}{2m\sqrt{k}} \).
03

Formulate the ODE

Apply Newton's 2nd Law to formulate the equation of motion for the damper-spring system: \(m\frac{d^2x}{dt^2} + 5\frac{dx}{dt} + kx = 0\). Here, we have an additional displacement of 4 inches or (1/3) feet when the weight is released.
04

Solve the ODE

Solve the second order ordinary differential equation (ODE) to describe the motion. The solution depends on the value of the damping ratio, \(\zeta\). There are three cases to consider: If \(\zeta<1\), the system is under-damped and oscillates. If \(\zeta>1\), the system is over-damped and will not oscillate but returns to equilibrium without oscillating. If \(\zeta=1\), the system is critically-damped and also does not oscillate.
05

Analyze the effect of the changed damping coefficient

If the damping coefficient was '4' instead of '5', we would have to recompute the damping ratio and reanalyze the system. Observe the effect this change has on the system's behavior, especially with regards to oscillation.

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