{"id":2025,"date":"2025-07-31T23:54:18","date_gmt":"2025-07-31T23:54:18","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2025"},"modified":"2026-03-23T16:12:00","modified_gmt":"2026-03-23T16:12:00","slug":"modeling-with-trigonometric-equations-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/modeling-with-trigonometric-equations-learn-it-3\/","title":{"raw":"Modeling with Trigonometric Equations: Learn It 3","rendered":"Modeling with Trigonometric Equations: Learn It 3"},"content":{"raw":"

Damped Harmonic Motion<\/h2>\r\nIn reality, a pendulum does not swing back and forth forever, nor does an object on a spring bounce up and down forever. Eventually, the pendulum stops swinging and the object stops bouncing and both return to equilibrium. Periodic motion in which an energy-dissipating force, or damping factor, acts is known as damped harmonic motion<\/strong>. Friction is typically the damping factor.\r\n\r\nIn physics, various formulas are used to account for the damping factor on the moving object. Some of these are calculus-based formulas that involve derivatives. For our purposes, we will use formulas for basic damped harmonic motion models.\r\n\r\n
\r\n

Damped Harmonic Motion<\/h3>\r\nIn damped harmonic motion<\/strong>, the displacement of an oscillating object from its rest position at time [latex]t[\/latex] is given as\r\n

[latex]f\\left(t\\right)=a{e}^{-ct}\\sin \\left(\\omega t\\right)\\text{or} f\\left(t\\right)=a{e}^{-ct}\\cos \\left(\\omega t\\right)[\/latex]<\/p>\r\nwhere [latex]c[\/latex] is a damping factor, [latex]|a|[\/latex] is the initial displacement and [latex]\\frac{2\\pi }{\\omega }[\/latex] is the period.\r\n\r\n<\/section>

\r\n

Model the equations that fit the two scenarios and use a graphing utility to graph the functions: Two mass-spring systems exhibit damped harmonic motion at a frequency of [latex]0.5[\/latex] cycles per second. Both have an initial displacement of 10 cm. The first has a damping factor of [latex]0.5[\/latex] and the second has a damping factor of [latex]0.1[\/latex].<\/span><\/h3>\r\n[reveal-answer q=\"574944\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"574944\"]\r\n\r\nAt time [latex]t=0[\/latex], the displacement is the maximum of 10 cm, which calls for the cosine function. The cosine function will apply to both models.\r\n\r\nWe are given the frequency [latex]f=\\frac{\\omega }{2\\pi }[\/latex] of 0.5 cycles per second. Thus,\r\n

[latex]\\begin{align}\\frac{\\omega }{2\\pi }&=0.5 \\\\ \\omega &=\\left(0.5\\right)2\\pi \\\\ &=\\pi \\end{align}[\/latex]<\/p>\r\nThe first spring system has a damping factor of [latex]c=0.5[\/latex]. Following the general model for damped harmonic motion, we have\r\n

[latex]f\\left(t\\right)=10{e}^{-0.5t}\\cos \\left(\\pi t\\right)[\/latex]<\/p>\r\n\"Graph\r\n\r\nThe second spring system has a damping factor of [latex]c=0.1[\/latex] and can be modeled as\r\n

[latex]f\\left(t\\right)=10{e}^{-0.1t}\\cos \\left(\\pi t\\right)[\/latex]<\/p>\r\n\"Graph\r\n

Analysis of the Solution<\/h4>\r\nNotice the differing effects of the damping constant. The local maximum and minimum values of the function with the damping factor [latex]c=0.5[\/latex] decreases much more rapidly than that of the function with [latex]c=0.1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
Find and graph a function of the form [latex]y=a{e}^{-ct}\\cos \\left(\\omega t\\right)[\/latex] that models the information given.\r\n
    \r\n \t
  1. [latex]a=20,c=0.05,p=4[\/latex]<\/li>\r\n \t
  2. [latex]a=2,c=1.5,f=3[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"923862\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"923862\"]\r\n\r\nSubstitute the given values into the model. Recall that period is [latex]\\frac{2\\pi }{\\omega }[\/latex] and frequency is [latex]\\frac{\\omega }{2\\pi }[\/latex].\r\n
      \r\n \t
    1. [latex]y=20{e}^{-0.05t}\\cos \\left(\\frac{\\pi }{2}t\\right)[\/latex].<\/li>\r\n \t
    2. [latex]y=2{e}^{-1.5t}\\cos \\left(6\\pi t\\right)[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
      The following equation represents a damped harmonic motion model: [latex]\\text{ }f\\left(t\\right)=5{e}^{-6t}\\cos \\left(4t\\right)[\/latex]\r\nFind the initial displacement, the damping constant, and the frequency.[reveal-answer q=\"533970\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"533970\"]initial displacement =6, damping constant = -6, frequency = [latex]\\frac{2}{\\pi }[\/latex][\/hidden-answer]<\/section>
      Find and graph a function of the form [latex]y=a{e}^{-ct}\\sin \\left(\\omega t\\right)[\/latex] that models the information given.\r\n
        \r\n \t
      1. [latex]a=7,c=10,p=\\frac{\\pi }{6}[\/latex]<\/li>\r\n \t
      2. [latex]a=0.3,c=0.2,f=20[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"293857\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"293857\"]\r\n\r\nCalculate the value of [latex]\\omega [\/latex] and substitute the known values into the model.\r\n
          \r\n \t
        1. As period is [latex]\\frac{2\\pi }{\\omega }[\/latex], we have\r\n
          [latex]\\begin{gathered}\\frac{\\pi }{6}=\\frac{2\\pi }{\\omega } \\\\ \\omega \\pi =6\\left(2\\pi \\right) \\\\ \\omega =12 \\end{gathered}[\/latex]<\/div>\r\nThe damping factor is given as 10 and the amplitude is 7. Thus, the model is [latex]y=7{e}^{-10t}\\sin \\left(12t\\right)[\/latex].<\/li>\r\n \t
        2. As frequency is [latex]\\frac{\\omega }{2\\pi }[\/latex], we have\r\n
          [latex]\\begin{gathered}20=\\frac{\\omega }{2\\pi } \\\\ 40\\pi =\\omega \\end{gathered}[\/latex]<\/div>\r\nThe damping factor is given as [latex]0.2[\/latex] and the amplitude is [latex]0.3[\/latex]. The model is [latex]y=0.3{e}^{-0.2t}\\sin \\left(40\\pi t\\right)[\/latex].<\/li>\r\n<\/ol>\r\n

          Analysis of the Solution<\/h4>\r\nA comparison of the last two examples illustrates how we choose between the sine or cosine functions to model sinusoidal criteria. We see that the cosine function is at the maximum displacement when [latex]t=0[\/latex], and the sine function is at the equilibrium point when [latex]t=0[\/latex]. For example, consider the equation [latex]y=20{e}^{-0.05t}\\cos \\left(\\frac{\\pi }{2}t\\right)[\/latex] from Example 9. We can see from the graph that when [latex]t=0,y=20[\/latex], which is the initial amplitude. Check this by setting [latex]t=0[\/latex] in the cosine equation:\r\n

          [latex]\\begin{align}y&=20{e}^{-0.05\\left(0\\right)}\\cos \\left(\\frac{\\pi }{2}\\cdot0\\right) \\\\ &=20\\left(1\\right)\\left(1\\right) \\\\ &=20\\end{align}[\/latex]<\/p>\r\nUsing the sine function yields\r\n

          [latex]\\begin{align}y&=20{e}^{-0.05\\left(0\\right)}\\sin \\left(\\frac{\\pi }{2}\\cdot0\\right) \\\\ &=20\\left(1\\right)\\left(0\\right) \\\\ &=0 \\end{align}[\/latex]<\/p>\r\nThus, cosine is the correct function.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

          Write the equation for damped harmonic motion given [latex]a=10,c=0.5[\/latex], and [latex]p=2[\/latex].[reveal-answer q=\"605857\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"605857\"][latex]y=10{e}^{-0.5t}\\cos \\left(\\pi t\\right)[\/latex][\/hidden-answer]<\/section>
          [ohm_question hide_question_numbers=1]173786[\/ohm_question]<\/section><\/div>","rendered":"

          Damped Harmonic Motion<\/h2>\n

          In reality, a pendulum does not swing back and forth forever, nor does an object on a spring bounce up and down forever. Eventually, the pendulum stops swinging and the object stops bouncing and both return to equilibrium. Periodic motion in which an energy-dissipating force, or damping factor, acts is known as damped harmonic motion<\/strong>. Friction is typically the damping factor.<\/p>\n

          In physics, various formulas are used to account for the damping factor on the moving object. Some of these are calculus-based formulas that involve derivatives. For our purposes, we will use formulas for basic damped harmonic motion models.<\/p>\n

          \n

          Damped Harmonic Motion<\/h3>\n

          In damped harmonic motion<\/strong>, the displacement of an oscillating object from its rest position at time [latex]t[\/latex] is given as<\/p>\n

          [latex]f\\left(t\\right)=a{e}^{-ct}\\sin \\left(\\omega t\\right)\\text{or} f\\left(t\\right)=a{e}^{-ct}\\cos \\left(\\omega t\\right)[\/latex]<\/p>\n

          where [latex]c[\/latex] is a damping factor, [latex]|a|[\/latex] is the initial displacement and [latex]\\frac{2\\pi }{\\omega }[\/latex] is the period.<\/p>\n<\/section>\n

          \n

          Model the equations that fit the two scenarios and use a graphing utility to graph the functions: Two mass-spring systems exhibit damped harmonic motion at a frequency of [latex]0.5[\/latex] cycles per second. Both have an initial displacement of 10 cm. The first has a damping factor of [latex]0.5[\/latex] and the second has a damping factor of [latex]0.1[\/latex].<\/span><\/h3>\n