A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The

Question

A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 12 ft/s. Determine the time at which the mass passes through the equilibrium position. (Use

g = 32 ft/s2

for the acceleration due to gravity.)
s

Find the time after the mass passes through the equilibrium position at which the mass attains its extreme displacement from the equilibrium position.
s

What is the position of the mass at this instant?
ft

A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 12 ft/s. Determine the time at which the mass passes through the equilibrium position. (Use g = 32 ft/s2 for the acceleration due to gravity.) Find the time after the mass passes through the equilibrium position at which the mass attains its extreme displacement from the equilibrium position What is the position of the mass at this instant? ft

Explanation / Answer

W = mg
4 = m*32 => m = 1/8 slug

The differential equation : (balance the force)

m* (d²x / dt²) = -kx - *(dx/dt) ; where ( = 1) is damping const., m is mass and k is spring constant
(1/8)(d²x / dt²) = -2x - 1*(dx/dt) ; or
d²x / dt² + 8 (dx/dt) + 16x = 0 ------> this is your diff eqn. of 2nd degree

D² + 8D + 16 = 0
(D + 4)² = 0
=> D1 = D2 = -4, Hence system is critically damped, and the general solution is
x(t) = C1 e^(-4t) + C2 t*e^(-4t)
Apply initial conditions, at t = 0, x(0) = -1 (initially 1 foot above equilibrium)
and initial velocity x'(0) = 12 ft/sec (down wards vel is positive)
-1 = C1 e^(0) + C2 *0*e^(0) => C1 = -1
and
x'(t) = C1 e^(-4t)*d/dt(-4t) + C2 [t*e^(-4t)*d/dt(-4t) + e^(-4t)*1]
12 = -1*e^(0)*(-4) + C2 [0*e^(0)*(-4) + e^(0)*1]
12 = 4 + C2 [0 + 1]
C2 = 8
x(t) = -e^(-4t) + 8 t*e^(-4t)

x(t)=0=-1+8t

t=1/8 sec

So x'(t) = 4 e^(-4t) + 8 e^(-4t) [1 - 4t]
we see that x'(t) = 0 = 4 e^(-4t) + 8 e^(-4t) [1 - 4t]
=>12 e^(-4t) = 32t e^(-4t)
=> t = 3/8 sec
and the corresponding extreme displacement can be found from x(t) = - e^(-4t) + 8t*e^(-4t)
x(t) = 8t*e^(-4t) - e^(-4t)
x(3/8) = 8*(3/8)*e^(-4*(3/8)) - e^(-4*(3/8))
x(3/8) = 3*e^(-3/2) - e^(-3/2)
x(3/8) = 2*e(-3/2)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.