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<h2 class="hd hd-2 unit-title">Introduction to Energy in SHM</h2>
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<p> So far we have seen Simple Harmonic Motion arise from a discussion of forces and Newton's Second Law. We can find the same characteristic differential equation describing Simple Harmonic Motion from an description of the energy as well.</p><p> Remeber that the spring force is a <b>conservative force</b>. The defining characteristic of a conservative force is that the work done moving the object between two points under the influence of that force depends only on those end points and not along the path taken. This allows us to define a potential energy associated with that force, as we have for the spring force, \(U= \frac{1}{2}kx^2\). In our model of the spring we are ignoring friction (a non-conservative force), and therefore we are dealing with a system that has only kinetic energy and potential energy; this allows us to write an energy conservation equation.</p><p> In this lesson we will see how to come back to our familiar SHM equations from an energy approach and how to generalize this approch to other systems with kinetic and potential energy.</p><p> Reading for this Lesson:</p><ul><li> Energy and Simple Harmonic Motion <a href="/courses/course-v1:MITx+8.01.4x+1T2019/pdfbook/0/chapter/23/12"><i class="it"> chapter 23.3 - 23.4</i></a></li></ul>
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<h2 class="hd hd-2 unit-title">L41v1: Energy Conservation for the Spring</h2>
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<h3 class="hd hd-2">L41v1: Energy Conservation for the Spring</h3>
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<h2 class="hd hd-2 unit-title">L41Q1: Motion in a Quadratic Potential</h2>
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Motion in a Quadratic Potential
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The potential energy function [mathjaxinline]U(x)=\frac{1}{2}kx^2[/mathjaxinline] for a particle with total mechanical energy [mathjaxinline]E[/mathjaxinline] is shown below. </p>
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<img src="/assets/courseware/v1/610cae083dcd6fc5c83a75c3f12b5dfd/asset-v1:MITx+8.01.4x+1T2019+type@asset+block/images_exam_final_cq_8.png" width="330"/>
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The particle was released moving in the positive [mathjaxinline]x[/mathjaxinline]-direction at the point [mathjaxinline]x=x_1[/mathjaxinline] at [mathjaxinline]t=0[/mathjaxinline]. The [mathjaxinline]x[/mathjaxinline]-component of the velocity of the particle at [mathjaxinline]t=0[/mathjaxinline] is </p>
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_exam_final_cq_8_2_1">
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<text> [mathjaxinline]v_ x(t=0)=0[/mathjaxinline],</text>
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<input type="radio" name="input_exam_final_cq_8_2_1" id="input_exam_final_cq_8_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="exam_final_cq_8_2_1-choice_2-label" for="input_exam_final_cq_8_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_exam_final_cq_8_2_1">
<text> [mathjaxinline]v_ x(t=0)=+\sqrt {\frac{2E}{m}}[/mathjaxinline],</text>
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<input type="radio" name="input_exam_final_cq_8_2_1" id="input_exam_final_cq_8_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="exam_final_cq_8_2_1-choice_3-label" for="input_exam_final_cq_8_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_exam_final_cq_8_2_1">
<text> [mathjaxinline]v_ x(t=0)=+\sqrt {\frac{2}{m}\left(E-\frac{1}{2}kx_1^2\right)}[/mathjaxinline],</text>
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<input type="radio" name="input_exam_final_cq_8_2_1" id="input_exam_final_cq_8_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="exam_final_cq_8_2_1-choice_4-label" for="input_exam_final_cq_8_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_exam_final_cq_8_2_1">
<text> [mathjaxinline]v_ x(t=0)=+\sqrt {\frac{2}{m}\left(\frac{1}{2}kx_1^2-E\right)}[/mathjaxinline],</text>
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<input type="radio" name="input_exam_final_cq_8_2_1" id="input_exam_final_cq_8_2_1_choice_5" class="field-input input-radio" value="choice_5"/><label id="exam_final_cq_8_2_1-choice_5-label" for="input_exam_final_cq_8_2_1_choice_5" class="response-label field-label label-inline" aria-describedby="status_exam_final_cq_8_2_1">
<text> [mathjaxinline]v_ x(t=0)=+\sqrt {\frac{kx_1^2}{m}}[/mathjaxinline],</text>
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<input type="radio" name="input_exam_final_cq_8_2_1" id="input_exam_final_cq_8_2_1_choice_6" class="field-input input-radio" value="choice_6"/><label id="exam_final_cq_8_2_1-choice_6-label" for="input_exam_final_cq_8_2_1_choice_6" class="response-label field-label label-inline" aria-describedby="status_exam_final_cq_8_2_1">
<text> [mathjaxinline]v_ x(t=0)=+\sqrt {-\frac{2E}{m}}[/mathjaxinline],</text>
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<div class="field">
<input type="radio" name="input_exam_final_cq_8_2_1" id="input_exam_final_cq_8_2_1_choice_7" class="field-input input-radio" value="choice_7"/><label id="exam_final_cq_8_2_1-choice_7-label" for="input_exam_final_cq_8_2_1_choice_7" class="response-label field-label label-inline" aria-describedby="status_exam_final_cq_8_2_1">
<text> [mathjaxinline]v_ x(t=0)=+\sqrt {-\frac{kx_1^2}{m}}[/mathjaxinline],</text>
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<h2 class="hd hd-2 unit-title">L41Q2: Spring Motion</h2>
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Spring Motion
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<p>A block of mass m = 1.0 kg is vibrating on a spring with a force constant of k = 298 N/m. When the instantaneous speed of the block is v = 0.7 m/s, the block is 0.30 m away from equilibrium.</p>
<br/>
<p>Find the total mechanical energy of the block-spring system during the vibrations.</p>
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<p>Find the maximum speed of the block during the vibrations.</p>
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<p>Find the amplitude of vibrations.</p>
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<h2 class="hd hd-2 unit-title">L41v2: Using Energy in Different Contexts</h2>
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<h2 class="hd hd-2 unit-title">L41v3: Derivition of the SHM Equation from Energy</h2>
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<center><p><b> Derivation of the simple harmonic oscillator equation using energy.</b></p></center><p><center><img src="/assets/courseware/v1/66f3df427c6f68f8366b09651140ae25/asset-v1:MITx+8.01.4x+1T2019+type@asset+block/images_html_L41_v01_intro.svg" width="350"/></center></p><p>A block of mass \(m\) is attached to the end of a horizontal spring of spring constant \(k\). The origin of the coordinate system is set at the equilibrium length of the spring. At time \(t\), the block is at \(x(t)\) and moving with a velocity \(v_x(t)\). Neglecting friction, the mechanical energy of the spring block system is constant. At time \(t\), the mechanical energy is given by:
</p><p>
\[ E = \dfrac{1}{2} m v_x^2 +\dfrac{1}{2}k x^2\]
</p><p>In the previous lessons we started with Newton's 2nd law and obtained the simple harmonic oscillation equation:
</p><p>
\[\Sigma \vec{\mathbf{F}} = m\vec{\mathbf{a}}\Longrightarrow \dfrac{d^2x}{dt^2} = -\dfrac{k}{m} x \].
</p><p>
In the video below this equation is derived using the fact that in the ideal block-spring system the mechanical energy is constant:
</p><p>
\[\dfrac{dE}{dt} = 0 \Longrightarrow \dfrac{d^2x}{dt^2} = -\dfrac{k}{m} x \]
</p><p><b>Note:</b> The kinetic energy, \(K = \dfrac{1}{2}mv_x^2(t)\), and the potential energy, \(U = \dfrac{1}{2}kx^2(t)\), are functions of time therefore:</p><p>
\[ \dfrac{dK}{dt}\neq 0 \mbox{ and } \dfrac{dU}{dt}\neq 0\]
</p><p>On the other hand, the mechanical energy is constant therefore:</p><p>
\[\dfrac{dE}{dt} = 0 = \dfrac{dK}{dt}+\dfrac{dU}{dt}\].</p>
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<h3 class="hd hd-2">L41v3: Derivition of the SHM Equation from Energy</h3>
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<h2 class="hd hd-2 unit-title">L41Q3: A Hanging Spring</h2>
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A hanging spring, part a
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A spring of spring constant [mathjaxinline]k[/mathjaxinline] and natural length [mathjaxinline]l_0[/mathjaxinline] is fixed to the ceiling as shown. When a block of mass [mathjaxinline]m[/mathjaxinline] is attached to the free end of the spring its new length is [mathjaxinline]l[/mathjaxinline]. </p>
<p><b class="bfseries">(Part a)</b> What is the value of [mathjaxinline]l[/mathjaxinline], the new equilibrium position? Express your answer in terms of [mathjaxinline]k[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]g[/mathjaxinline], and l_0 for [mathjaxinline]l_0[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]l=[/mathjaxinline] </p>
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A hanging spring, part b
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<p><b class="bfseries">(Part b)</b> Consider the coordinate system shown in the figure below where [mathjaxinline]y=0[/mathjaxinline] is set at the ceiling. </p>
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At a given instant of time [mathjaxinline]t[/mathjaxinline], the box is at position [mathjaxinline]y(t)[/mathjaxinline] from the ceiling, and moving with [mathjaxinline]v_ y(t)[/mathjaxinline]. What is [mathjaxinline]U_ s[/mathjaxinline], the elastic potential energy? Consider the zero of potential energy at [mathjaxinline]y=l_0[/mathjaxinline]. Express your answer in terms of [mathjaxinline]l[/mathjaxinline], l_0 for [mathjaxinline]l_0[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], and [mathjaxinline]y[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]U_ s=[/mathjaxinline] </p>
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A hanging spring, part c
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<p><b class="bfseries">(Part c)</b> At the instant of time shown in the figure presented in part (b), what is [mathjaxinline]U_ g[/mathjaxinline], the gravitational potential energy? Consider the zero of potential energy at [mathjaxinline]y=l_0[/mathjaxinline]. Express your answer in terms of [mathjaxinline]l[/mathjaxinline], l_0 for [mathjaxinline]l_0[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]g[/mathjaxinline], and [mathjaxinline]y[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]U_ g=[/mathjaxinline] </p>
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A hanging spring, part d
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<p><b class="bfseries">(Part d)</b> Use the results of parts (b) and (c) to write [mathjaxinline]U = U_ s+U_ g[/mathjaxinline], the total potential energy of the block at the instant when it is at [mathjaxinline]y(t)[/mathjaxinline] from the ceiling. What is [mathjaxinline]\dfrac {dU}{dt}[/mathjaxinline], the time derivative of [mathjaxinline]U[/mathjaxinline]. Express your answer in terms of l_0 for [mathjaxinline]l_0[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]g[/mathjaxinline], [mathjaxinline]y[/mathjaxinline], v_y for [mathjaxinline]v_ y[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]\dfrac {dU}{dt} =[/mathjaxinline] </p>
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A hanging spring, part e
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<p><b class="bfseries">(Part e)</b> At the instant shown in the figure presented in part (b), the box's kinetic energy is given by [mathjaxinline]K = \dfrac {1}{2}mv_ y^2[/mathjaxinline]. Let [mathjaxinline]a_ y[/mathjaxinline] be the box's acceleration: [mathjaxinline]a_ y =\dfrac {dv_ y}{dt}[/mathjaxinline]. Calculate [mathjaxinline]\dfrac {dK}{dt}[/mathjaxinline], the time derivative of the box's kinetic energy. Express your answer in terms of [mathjaxinline]m[/mathjaxinline], v_y for [mathjaxinline]v_ y[/mathjaxinline], a_y for [mathjaxinline]a_ y[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\dfrac {dK}{dt} =[/mathjaxinline] </p>
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A hanging spring, part f
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<p><b class="bfseries">(Part f)</b> Use the results of parts (d) and (e) to find [mathjaxinline]\dfrac {dE}{dt}[/mathjaxinline], the time derivative of the mechanical energy [mathjaxinline]E= K+U_ s+U_ g[/mathjaxinline]. Express your answer in terms of l_0 for [mathjaxinline]l_0[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]g[/mathjaxinline], [mathjaxinline]y[/mathjaxinline], v_y for [mathjaxinline]v_ y[/mathjaxinline], and a_y for [mathjaxinline]a_ y[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]\dfrac {dE}{dt} =[/mathjaxinline] </p>
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A hanging spring, part g
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<p><b class="bfseries">(Part g)</b> Use the results of part (f) and the fact that [mathjaxinline]\dfrac {dE}{dt} = 0[/mathjaxinline] to obtain an expression of [mathjaxinline]a_ y = \dfrac {d^2y}{dt^2}[/mathjaxinline]. Express your answer in terms of l_0 for [mathjaxinline]l_0[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]g[/mathjaxinline], [mathjaxinline]y[/mathjaxinline], and v_y for [mathjaxinline]v_ y[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]\dfrac {d^2y}{dt^2} =[/mathjaxinline] </p>
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A hanging spring, part h
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<p><b class="bfseries">(Part h)</b> Use the result of part (a) to write the equation of motion obtained in part (g) in terms of [mathjaxinline]l[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], and [mathjaxinline]y[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\dfrac {d^2y}{dt^2} =[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">L41v4: Generalized Energy in SHM</h2>
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<h2 class="hd hd-2 unit-title">L41v5: Worked Example - Rolling Cylinder on a Spring</h2>
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Worked Example: Rolling Cylinder on a Spring
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Attach a solid cylinder of mass [mathjaxinline]M[/mathjaxinline] and radius [mathjaxinline]R[/mathjaxinline] to a horizontal massless spring with spring constant [mathjaxinline]k[/mathjaxinline] so that it can roll without slipping along a horizontal surface. The moment of inertia of the cylinder about the center of mass is [mathjaxinline]I_{cm}=(1/2)MR^2[/mathjaxinline]. At time [mathjaxinline]t[/mathjaxinline], the center of mass of the cylinder is moving with velocity [mathjaxinline]\vec V_{cm}=V_{cm}\hat i[/mathjaxinline], rotating with angular velocity [mathjaxinline]\vec\omega =\omega \hat k[/mathjaxinline] and the spring is compressed a distance [mathjaxinline]x[/mathjaxinline] from its equilibrium length. What is the period of simple harmonic motion for the center of mass of the cylinder? </p>
<p><b class="bfseries">(Part a))</b> At time [mathjaxinline]t[/mathjaxinline], what is the relationship between the center of mass speed and the angular speed? Write your answer using some or all of the following: [mathjaxinline]M[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline]. </p>
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<p><b class="bfseries">(Part b))</b> What is the mechanical energy [mathjaxinline]E[/mathjaxinline] of the cylinder-spring system at time [mathjaxinline]t[/mathjaxinline]? Write your answer using some or all of the following: [mathjaxinline]M[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], [mathjaxinline]x[/mathjaxinline], and V for [mathjaxinline]V_{cm}[/mathjaxinline]. </p>
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<p><b class="bfseries">(Part c))</b> At time [mathjaxinline]t[/mathjaxinline], use the condition that the mechanical energy is constant and therefore [mathjaxinline]\frac{dE}{dt}=0[/mathjaxinline], to determine a differential equation satisfied by the function [mathjaxinline]x(t)[/mathjaxinline]. Write your answer using some or all of the following: [mathjaxinline]R[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], [mathjaxinline]x[/mathjaxinline], and [mathjaxinline]M[/mathjaxinline]. </p>
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<p><b class="bfseries">(Part d))</b> What is the angular frequency [mathjaxinline]\omega _0[/mathjaxinline] associated with the simple harmonic motion of the center of mass of the cylinder? Write your answer using some or all of the following: [mathjaxinline]R[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], [mathjaxinline]x[/mathjaxinline], and [mathjaxinline]M[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\omega _0=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part e))</b> What is the period [mathjaxinline]T[/mathjaxinline] associated with the simple harmonic motion of the center of mass of the cylinder? Write your answer using some or all of the following: pi for [mathjaxinline]\pi[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], [mathjaxinline]x[/mathjaxinline], and [mathjaxinline]M[/mathjaxinline]. </p>
<p>
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<h3 class="hd hd-2">L41v5: Worked Example - Rolling Cylinder on a Spring</h3>
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<h2 class="hd hd-2 unit-title">L41v6: Worked Example - U-tube</h2>
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<h3 class="hd hd-2">L41v6: Worked Example - U-tube</h3>
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<h3 class="hd hd-4 downloads-heading sr" id="video-download-transcripts_L41v06">Downloads and transcripts</h3>
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<h4 class="hd hd-5">Transcripts</h4>
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<h2 class="hd hd-2 unit-title">Deep Dive - Integrate Energy to Find the Simple Harmonic Motion Solution</h2>
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<h3 class="hd hd-2">L41: Deep Dive - Integrate Energy to Find the Simple Harmonic Motion Solution</h3>
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