College Physics for AP® Courses

# 16.5Energy and the Simple Harmonic Oscillator

College Physics for AP® Courses16.5 Energy and the Simple Harmonic Oscillator

### Learning Objectives

By the end of this section, you will be able to:

• Describe the changes in energy that occur while a system undergoes simple harmonic motion.

To study the energy of a simple harmonic oscillator, we first consider all the forms of energy it can have We know from Hooke’s Law: Stress and Strain Revisited that the energy stored in the deformation of a simple harmonic oscillator is a form of potential energy given by:

$PE el = 1 2 kx 2 . PE el = 1 2 kx 2 . size 12{"PE" size 8{"el"}= { {1} over {2} } ital "kx" rSup { size 8{2} } } {}$
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Because a simple harmonic oscillator has no dissipative forces, the other important form of energy is kinetic energy $KEKE size 12{ ital "KE"} {}$. Conservation of energy for these two forms is:

$KE + PE el = constant KE + PE el = constant size 12{ ital "KE"+ ital "PE" rSub { size 8{e1} } ="constant"} {}$
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or

$1 2 mv 2 + 1 2 kx 2 = constant. 1 2 mv 2 + 1 2 kx 2 = constant. size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } + { {1} over {2} } ital "kx" rSup { size 8{2} } ="constant"} {}$
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This statement of conservation of energy is valid for all simple harmonic oscillators, including ones where the gravitational force plays a role

Namely, for a simple pendulum we replace the velocity with $v=Lωv=Lω size 12{v=Lω} {}$, the spring constant with $k=mg/Lk=mg/L size 12{k= ital "mg"/L} {}$, and the displacement term with $x=Lθx=Lθ size 12{x=Lθ} {}$. Thus

$1 2 mL 2 ω 2 + 1 2 mgL θ 2 = constant. 1 2 mL 2 ω 2 + 1 2 mgL θ 2 = constant. size 12{ { {1} over {2} } ital "mL" rSup { size 8{2} } ω rSup { size 8{2} } + { {1} over {2} } ital "mgL"θ rSup { size 8{2} } ="constant"} {}$
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In the case of undamped simple harmonic motion, the energy oscillates back and forth between kinetic and potential, going completely from one to the other as the system oscillates. So for the simple example of an object on a frictionless surface attached to a spring, as shown again in Figure 16.16, the motion starts with all of the energy stored in the spring. As the object starts to move, the elastic potential energy is converted to kinetic energy, becoming entirely kinetic energy at the equilibrium position. It is then converted back into elastic potential energy by the spring, the velocity becomes zero when the kinetic energy is completely converted, and so on. This concept provides extra insight here and in later applications of simple harmonic motion, such as alternating current circuits.

Figure 16.16 The transformation of energy in simple harmonic motion is illustrated for an object attached to a spring on a frictionless surface.

The conservation of energy principle can be used to derive an expression for velocity $vv size 12{v} {}$. If we start our simple harmonic motion with zero velocity and maximum displacement ($x=Xx=X size 12{x=X} {}$), then the total energy is

$12kX2.12kX2. size 12{ { {1} over {2} } ital "kX" rSup { size 8{2} } } {}$
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This total energy is constant and is shifted back and forth between kinetic energy and potential energy, at most times being shared by each. The conservation of energy for this system in equation form is thus:

$1 2 mv 2 + 1 2 kx 2 = 1 2 kX 2 . 1 2 mv 2 + 1 2 kx 2 = 1 2 kX 2 . size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } + { {1} over {2} } ital "kx" rSup { size 8{2} } = { {1} over {2} } ital "kX" rSup { size 8{2} } } {}$
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Solving this equation for $vv size 12{v} {}$ yields:

$v = ± k m X 2 − x 2 . v = ± k m X 2 − x 2 . size 12{v= +- sqrt { { {k} over {m} } left (X rSup { size 8{2} } - x rSup { size 8{2} } right )} } {}$
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Manipulating this expression algebraically gives:

$v = ± k m X 1 − x 2 X 2 v = ± k m X 1 − x 2 X 2 size 12{v= +- sqrt { { {k} over {m} } } X sqrt {1 - { {x rSup { size 8{2} } } over {X rSup { size 8{2} } } } } } {}$
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and so

$v = ± v max 1 − x 2 X 2 , v = ± v max 1 − x 2 X 2 , size 12{v= +- v size 8{"max" sqrt {1 - { {x rSup { size 8{2} } } over {X rSup { size 8{2} } } } } }} {}$
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where

$v max = k m X . v max = k m X . size 12{v size 8{"max"}= sqrt { { {k} over {m} } } X} {}$
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From this expression, we see that the velocity is a maximum ($vmaxvmax$) at $x=0x=0 size 12{x=0} {}$, as stated earlier in $vt=−vmaxsin2πtTvt=−vmaxsin2πtT$. Notice that the maximum velocity depends on three factors. Maximum velocity is directly proportional to amplitude. As you might guess, the greater the maximum displacement the greater the maximum velocity. Maximum velocity is also greater for stiffer systems, because they exert greater force for the same displacement. This observation is seen in the expression for $vmax;vmax;$ it is proportional to the square root of the force constant $kk$. Finally, the maximum velocity is smaller for objects that have larger masses, because the maximum velocity is inversely proportional to the square root of $mm$. For a given force, objects that have large masses accelerate more slowly.

A similar calculation for the simple pendulum produces a similar result, namely:

$ω max = g L θ max . ω max = g L θ max . size 12{ω rSub { size 8{"max"} } = sqrt { { {g} over {L} } } θ rSub { size 8{"max"} } } {}$
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### Making Connections: Mass Attached to a Spring

Consider a mass m attached to a spring, with spring constant k, fixed to a wall. When the mass is displaced from its equilibrium position and released, the mass undergoes simple harmonic motion. The spring exerts a force $F = −kv F = −kv$ on the mass. The potential energy of the system is stored in the spring. It will be zero when the spring is in the equilibrium position. All the internal energy exists in the form of kinetic energy, given by $KE = 1 2 m v 2 KE = 1 2 m v 2$. As the system oscillates, which means that the spring compresses and expands, there is a change in the structure of the system and a corresponding change in its internal energy. Its kinetic energy is converted to potential energy and vice versa. This occurs at an equal rate, which means that a loss of kinetic energy yields a gain in potential energy, thus preserving the work-energy theorem and the law of conservation of energy.

### Example 16.7

#### Determine the Maximum Speed of an Oscillating System: A Bumpy Road

Suppose that a car is 900 kg and has a suspension system that has a force constant $k=6.53×104N/mk=6.53×104N/m size 12{k=6 "." "53" times "10" rSup { size 8{4} } `"N/m"} {}$. The car hits a bump and bounces with an amplitude of 0.100 m. What is its maximum vertical velocity if you assume no damping occurs?

#### Strategy

We can use the expression for $vmaxvmax size 12{v rSub { size 8{"max"} } } {}$ given in $vmax=kmXvmax=kmX size 12{v size 8{"max"}= sqrt { { {k} over {m} } } X} {}$ to determine the maximum vertical velocity. The variables $mm size 12{m} {}$ and $kk size 12{k} {}$ are given in the problem statement, and the maximum displacement $XX size 12{X} {}$ is 0.100 m.

#### Solution

1. Identify known.
2. Substitute known values into $vmax=kmXvmax=kmX size 12{v size 8{"max"}= sqrt { { {k} over {m} } } X} {}$:
$v max = 6 . 53 × 10 4 N/m 900 kg (0 . 100 m) . v max = 6 . 53 × 10 4 N/m 900 kg (0 . 100 m) . size 12{v size 8{"max"}= sqrt { { {6 "." "53" times "10" rSup { size 8{4} } "N/m"} over {"900"" kg"} } } 0 "." "100"" m"} {}$
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3. Calculate to find $vmax= 0.852 m/s.vmax= 0.852 m/s. size 12{v rSub { size 8{"max"} } } {}$

#### Discussion

This answer seems reasonable for a bouncing car. There are other ways to use conservation of energy to find $vmaxvmax size 12{v rSub { size 8{"max"} } } {}$. We could use it directly, as was done in the example featured in Hooke’s Law: Stress and Strain Revisited.

The small vertical displacement $yy size 12{v rSub { size 8{"max"} } } {}$ of an oscillating simple pendulum, starting from its equilibrium position, is given as

$y ( t ) = a sin ωt , y ( t ) = a sin ωt , size 12{y $$t$$ =a"sin"ωt} {}$
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where $aa size 12{a} {}$ is the amplitude, $ωω size 12{ω} {}$ is the angular velocity and $tt size 12{t} {}$ is the time taken. Substituting $ω=2πTω=2πT size 12{ω= { {2π} over {T} } } {}$, we have

$y t = a sin 2πt T . y t = a sin 2πt T . size 12{y left (t right )=a"sin" left ( { {2πt} over {T} } right )} {}$
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Thus, the displacement of pendulum is a function of time as shown above.

Also the velocity of the pendulum is given by

$v ( t ) = 2aπ T cos 2πt T , v ( t ) = 2aπ T cos 2πt T , size 12{v $$t$$ = { {2aπ} over {T} } "cos" left ( { {2πt} over {T} } right )} {}$
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so the motion of the pendulum is a function of time.

Why does it hurt more if your hand is snapped with a ruler than with a loose spring, even if the displacement of each system is equal?

You are observing a simple harmonic oscillator. Identify one way you could decrease the maximum velocity of the system.