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A 2-kg mass is attached to a spring with spring constant 24 N/m. The system is then immersed in a medium imparting a damping force equal to 16 times the instantaneous velocity of the mass. Find the equation of motion if it is released from rest at a point 40 cm below equilibrium.

x ( t ) = 0.6 e −2 t 0.2 e −6 t

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Case 2: b 2 = 4 m k

In this case, we say the system is critically damped . The general solution has the form

x ( t ) = c 1 e λ 1 t + c 2 t e λ 1 t ,

where λ 1 is less than zero. The motion of a critically damped system is very similar to that of an overdamped system. It does not oscillate. However, with a critically damped system, if the damping is reduced even a little, oscillatory behavior results. From a practical perspective, physical systems are almost always either overdamped or underdamped (case 3, which we consider next). It is impossible to fine-tune the characteristics of a physical system so that b 2 and 4 m k are exactly equal. [link] shows what typical critically damped behavior looks like.

This figure has two graphs labeled (a) and (b). The first graph is in the first quadrant and is a decreasing curve with the horizontal axis as a horizontal asymptote. The second graph initially is a decreasing function but becomes increasing below the horizontal axis. Then, the horizontal axis is also a horizontal asymptote.
Behavior of a critically damped spring-mass system. The system graphed in part (a) has more damping than the system graphed in part (b).

Critically damped spring-mass system

A 1-kg mass stretches a spring 20 cm. The system is attached to a dashpot that imparts a damping force equal to 14 times the instantaneous velocity of the mass. Find the equation of motion if the mass is released from equilibrium with an upward velocity of 3 m/sec.

We have m g = 1 ( 9.8 ) = 0.2 k , so k = 49 . Then, the differential equation is

x + 14 x + 49 x = 0 ,

which has general solution

x ( t ) = c 1 e −7 t + c 2 t e −7 t .

Applying the initial conditions x ( 0 ) = 0 and x ( 0 ) = −3 gives

x ( t ) = −3 t e −7 t .
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A 1-lb weight stretches a spring 6 in., and the system is attached to a dashpot that imparts a damping force equal to half the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point 6 in. below equilibrium.

x ( t ) = 1 2 e −8 t + 4 t e −8 t

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Case 3: b 2 < 4 m k

In this case, we say the system is underdamped . The general solution has the form

x ( t ) = e α t ( c 1 cos ( β t ) + c 2 sin ( β t ) ) ,

where α is less than zero. Underdamped systems do oscillate because of the sine and cosine terms in the solution. However, the exponential term dominates eventually, so the amplitude of the oscillations decreases over time. [link] shows what typical underdamped behavior looks like.

This figure is an oscillating graph where the amplitude is decreasing. There are red dashed curves at the peaks of the amplitudes showing the pattern of a decreasing amplitude. As t increases, the horizontal axis becomes a horizontal asymptote.
Behavior of an underdamped spring-mass system.

Note that for all damped systems, lim t x ( t ) = 0 . The system always approaches the equilibrium position over time.

Underdamped spring-mass system

A 16-lb weight stretches a spring 3.2 ft. Assume the damping force on the system is equal to the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point 9 in. below equilibrium.

We have k = 16 3.2 = 5 and m = 16 32 = 1 2 , so the differential equation is

1 2 x + x + 5 x = 0 , or x + 2 x + 10 x = 0 .

This equation has the general solution

x ( t ) = e t ( c 1 cos ( 3 t ) + c 2 sin ( 3 t ) ) .

Applying the initial conditions, x ( 0 ) = 3 4 and x ( 0 ) = 0 , we get

x ( t ) = e t ( 3 4 cos ( 3 t ) + 1 4 sin ( 3 t ) ) .
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A 1-kg mass stretches a spring 49 cm. The system is immersed in a medium that imparts a damping force equal to four times the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point 24 cm above equilibrium.

x ( t ) = −0.24 e −2 t cos ( 4 t ) 0.12 e −2 t sin ( 4 t )

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Questions & Answers

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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