Worksheet: Modeling with Second-Order Differential Equations

In this worksheet, we will practice using second-order differential equations to model situations in real life.

Q1:

Consider a body of mass ๐‘š attached to a spring with spring constant ๐‘˜. The differential equation ๐‘š๐‘ฆ๐‘ก=โˆ’๐‘˜๐‘ฆdd๏Šจ๏Šจ, where ๐‘ฆ is the vertical displacement, can be used to model the dynamics of this system. However, such a differential equation implies that, once the body starts moving, it will oscillate forever. A better model is one that also considers the effects of frictional forces. If we add a frictional force that is proportional to the velocity of the body, we get the following differential equation ๐‘š๐‘ฆ๐‘ก=โˆ’๐‘˜๐‘ฆโˆ’๐‘ ๐‘ฆ๐‘ก,dddd๏Šจ๏Šจ where ๐‘ >0.

Letting ๐‘ค=๐‘˜๐‘š๏Šจ and 2๐‘=๐‘ ๐‘š, there are three possible behaviours of the solution to this equation which we can define in terms of ๐‘๏Šจ and ๐‘ค๏Šจ as follows: over damped when ๐‘>๐‘ค๏Šจ๏Šจ; critically damped when ๐‘=๐‘ค๏Šจ๏Šจ; and underdamped or oscillatory when ๐‘<๐‘ค๏Šจ๏Šจ. Which of the following solutions can be used to describe one of these behaviours:

  1. ๐‘ฆ=(๐‘+๐‘๐‘ก)๐‘’๏Šง๏Šจ๏Šฑ๏Œป๏
  2. ๐‘ฆ=๐‘๐‘’+๐‘๐‘’๏Šง๏Šฑ๏ฟ๏๏Šจ๏Šฑ๏Ž€๏, where ๐œ†=๐‘+โˆš๐‘โˆ’๐‘ค๏Šจ๏Šจ and ๐œ‡=๐‘โˆ’โˆš๐‘โˆ’๐‘ค๏Šจ๏Šจ
  3. ๐‘ฆ=๐‘๐‘’(๐›ฝ๐‘ก+๐›พ)๏Šฑ๏Œป๏sin where ๐›ฝ=โˆš๐‘คโˆ’๐‘๏Šจ๏Šจ
  • Ab
  • Ba
  • CEach solution describes one of the different behaviours
  • Dc

Q2:

The one-dimensional time-independent Schrodinger equation is given as dd๏Šจ๏Šจ๐œ“๐‘ฅ=2๐‘šโ„[๐‘ˆ(๐‘ฅ)โˆ’๐ธ]๐œ“,

where ๐œ“ is a wave function which describes the displacement ๐‘ฅ of a single particle of mass ๐‘š, ๐ธ is the total energy, ๐‘ˆ is the potential energy, and โ„ is a known constant. Since ๐‘ˆ(๐‘ฅ)=0 for the particle-in-a-box model, where 0โ‰ค๐‘ฅโ‰ค๐‘Ž, this second-order differential equation becomes ๐œ“โ€ฒโ€ฒ=๐›ผ๐œ“,๐›ผ=โˆ’2๐‘š๐ธโ„.๏Šจ๏Šจ๏Šจwhere

Find the general solution for this differential equation.

  • A๐‘ฆ=(๐‘+๐‘)๐‘’๏Šง๏Šจ๏ต๏—
  • B๐‘ฆ=๐‘๐‘’+๐‘๐‘’๏Šง๏ต๏—๏Šจ๏Šฑ๏ต๏—
  • C๐‘ฆ=๐‘๐‘’+๐‘๐‘’๏Šง๏Šจ๏‘ฎ๏‘๏Žช๏‘ฎ๏‘
  • D๐‘ฆ=(๐‘+๐‘)๐‘’๏Šง๏Šจ๏Šฑ๏ต๏—

Q3:

Consider a mass ๐‘š that oscillates at the end of a spring having a spring constant ๐‘˜. The following second-order differential equation describes the vertical displacement ๐‘ฆ of this spring-mass system: ๐‘š๐‘ฆ๐‘ฅ=โˆ’๐‘˜๐‘ฆ.dd๏Šจ๏Šจ This differential equation neglects the influence of air resistance or frictional forces. Find the general solution which describes the vertical displacement of this spring as a function of time ๐‘ก.

  • A๐‘ฆ=๐‘๏€ผ๐‘ก๐œ”๏ˆ+๐‘๏€ผ๐‘ก๐œ”๏ˆ๏Šง๏Šจsincos, ๐œ”=๐‘˜๐‘š๏Šจ
  • B๐‘ฆ=๐‘๏€ผ๐‘ก๐œ”๏ˆ+๐‘๏€ผ๐‘ก๐œ”๏ˆ๏Šง๏Šจ๏Šจ๏Šจsincos, ๐œ”=๐‘˜๐‘š๏Šจ
  • C๐‘ฆ=๐‘(๐œ”๐‘ก)+๐‘(๐œ”๐‘ก)๏Šง๏Šจsincos, ๐œ”=๐‘˜๐‘š๏Šจ
  • D๐‘ฆ=๐‘(๐œ”๐‘ก)+๐‘(๐œ”๐‘ก)๏Šง๏Šจ๏Šจ๏Šจsincos, ๐œ”=๐‘˜๐‘š๏Šจ

Q4:

Consider a mass ๐‘š that oscillates at the end of a spring having a spring constant ๐‘˜. The following second-order differential equation describes the vertical displacement ๐‘ฆ of this spring-mass system: ๐‘š๐‘ฆ๐‘ฅ=โˆ’๐‘˜๐‘ฆ.dd๏Šจ๏Šจ

Such a differential equation implies that mass ๐‘š, once started, will simply oscillate up and down forever! This differential equation neglects the influence of frictional forces. Let us assume that there is a retarding force which is proportional to the velocity of motion where ๐‘  is treated as a proportionality constant (๐‘ >0). The aforementioned differential equation now becomes ๐‘š๐‘ฆ๐‘ฅ=โˆ’๐‘˜๐‘ฆโˆ’๐‘ ๐‘ฆ๐‘ฅdddd๏Šจ๏Šจ

Let ๐œ”=๐‘˜๐‘š๏Šจ and 2๐‘=๐‘ ๐‘š, where ๐‘>0.

There are three possible types of solutions which depend upon the relative size of ๐‘๏Šจ and ๐œ”๏Šจ, including the following:

overdamped if ๐‘>๐œ”๏Šจ๏Šจ,

critically damped if ๐‘=๐œ”๏Šจ๏Šจ,

underdamped or oscillatory if ๐‘<๐œ”๏Šจ๏Šจ.

Find the general solution for this differential equation, which describes the vertical displacement of this spring as a function of time ๐‘ก for โ€œoverdampedโ€ motion.

  • A๐‘ฆ=๐‘๐‘ก๐‘’+๐‘๐‘’๏Šง๏ฟ๏๏Šจ๏Ž€๏, ๐œ†=๐‘+โˆš๐‘โˆ’๐œ”๏Šจ๏Šจ, ๐œ‡=๐‘โˆ’โˆš๐‘โˆ’๐œ”๏Šจ๏Šจ
  • B๐‘ฆ=๐‘๐‘ก๐‘’+๐‘๐‘’๏Šง๏Šฑ๏Šจ๏Šฑ๏‘‰๏‘ธ๏‘‰๏‘น, ๐œ†=๐‘+โˆš๐‘โˆ’๐œ”๏Šจ๏Šจ, ๐œ‡=๐‘โˆ’โˆš๐‘โˆ’๐œ”๏Šจ๏Šจ
  • C๐‘ฆ=๐‘๐‘’+๐‘๐‘’๏Šง๏Šฑ๏ฟ๏๏Šจ๏Šฑ๏Ž€๏, ๐œ†=๐‘+โˆš๐‘โˆ’๐œ”๏Šจ๏Šจ, ๐œ‡=๐‘โˆ’โˆš๐‘โˆ’๐œ”๏Šจ๏Šจ
  • D๐‘ฆ=๐‘๐‘’+๐‘๐‘’๏Šง๏Šจ๏‘‰๏‘ธ๏‘‰๏‘น, ๐œ†=๐‘+โˆš๐‘โˆ’๐œ”๏Šจ๏Šจ, ๐œ‡=๐‘โˆ’โˆš๐‘โˆ’๐œ”๏Šจ๏Šจ

Q5:

Consider a mass ๐‘š that oscillates at the end of a spring having a spring constant ๐‘˜. The second-order differential equation ๐‘š๐‘‘๐‘ฆ๐‘‘๐‘ฅ=โˆ’๐‘˜๐‘ฆ๏Šจ๏Šจ describes the vertical displacement ๐‘ฆ of this springโ€“mass system.

Such a differential equation implies that mass ๐‘š, once started, will simply oscillate up and down forever. This differential equation neglects the influence of frictional forces. Let us assume that there is a retarding force that is proportional to the velocity of motion where ๐‘  is treated as a proportionality constant (๐‘ >0). The aforementioned differential equation now becomes ๐‘š๐‘‘๐‘ฆ๐‘‘๐‘ก=โˆ’๐‘˜๐‘ฆโˆ’๐‘ ๐‘ฆ๐‘ก.๏Šจ๏Šจdd

Let ๐œ”=๐‘˜๐‘š๏Šจ and 2๐‘=๐‘ ๐‘š where ๐‘>0.

There are three possible types of solutions that depend on the relative size of ๐‘๏Šจ and ๐œ”๏Šจ, including the following:

overdamped if ๐‘>๐œ”๏Šจ๏Šจ,

critically damped if ๐‘=๐œ”๏Šจ๏Šจ,

underdamped or oscillatory if ๐‘<๐œ”๏Šจ๏Šจ.

Find the general solution for this differential equation, which describes the vertical displacement of this spring as a function of time ๐‘ก for the โ€œcritically dampedโ€ motion."

  • A๐‘ฆ=(๐‘+๐‘๐‘ก)๐‘’๏Šง๏Šจ๏Šฑ๏Œป๏
  • B๐‘ฆ=๐‘๐‘’+๐‘๐‘’๏Šง๏Šฑ๏Œป๏๏Šจ๏Œป๏
  • C๐‘ฆ=(๐‘+๐‘)๐‘’๏Šง๏Šจ๏Œป๏
  • D๐‘ฆ=๐‘๐‘’+๐‘๐‘’๏Šง๏Šฑ๏Šจ๏‘‰๏ด๏‘‰๏ด

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