Question

a 0.75-kg mass is attached to the lower end of a coil spring suspended from the ceiling and having a spring constant of 48 n/m. the resistance in the spring-mass system is numerically equal to the instantaneous velocity. at t = 0 the mass is set in motion from a position 0.25 m below its equilibrium position by giving it a downward velocity of 0.25 m/s. write an initial value problem that models the given situation.

write the differential equation for the motion of the mass. the positive direction is downward.

y + y + y = \square
(simplify your answer. use integers or decimals for any numbers in the expression.)

Explanation:

Step1: Find the mass

Given \(m = 0.15\space kg\).

Step2: Determine the damping coefficient \(β\)

Since the resistance is numerically equal to the instantaneous velocity, \(β = 1\).

Step3: Find the spring constant \(k\)

Given \(k = 48\space N/m\).

Step4: Write the differential equation

Using the formula \(my''+βy'+ky = 0\), substitute \(m = 0.15\), \(β = 1\), and \(k = 48\).
We get \(0.15y''+y'+48y = 0\).

Step5: Write the initial conditions

At \(t = 0\), \(y(0)=\frac{6}{12}=\frac{1}{2}\space m\) (since the mass is \(\frac{6}{12}m\) below equilibrium) and \(y'(0)=0.25\space m/s\) (downward velocity).

Answer:

The differential equation is \(0.15y'' + y'+48y = 0\) with initial conditions \(y(0)=\frac{1}{2}\) and \(y'(0) = 0.25\).