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m x + k x = 0 .

It is convenient to rearrange this equation and introduce a new variable, called the angular frequency , ω . Letting ω = k / m , we can write the equation as

x + ω 2 x = 0 .

This differential equation has the general solution

x ( t ) = c 1 cos ω t + c 2 sin ω t ,

which gives the position of the mass at any point in time. The motion of the mass is called simple harmonic motion    . The period of this motion (the time it takes to complete one oscillation) is T = 2 π ω and the frequency is f = 1 T = ω 2 π ( [link] ).

This figure is the graph of f(t) = sin 2t. It is a periodic, oscillating graph. The period of the graph is represented with a line pointing from one peak to the next. It is labeled with the period  T = 2π/ω.
A graph of vertical displacement versus time for simple harmonic motion.

Simple harmonic motion

Assume an object weighing 2 lb stretches a spring 6 in. Find the equation of motion if the spring is released from the equilibrium position with an upward velocity of 16 ft/sec. What is the period of the motion?

We first need to find the spring constant. We have

m g = k s 2 = k ( 1 2 ) k = 4 .

We also know that weight W equals the product of mass m and the acceleration due to gravity g . In English units, the acceleration due to gravity is 32 ft/sec 2 .

W = m g 2 = m ( 32 ) m = 1 16

Thus, the differential equation representing this system is

1 16 x + 4 x = 0 .

Multiplying through by 16, we get x + 64 x = 0 , which can also be written in the form x + ( 8 2 ) x = 0 . This equation has the general solution

x ( t ) = c 1 cos ( 8 t ) + c 2 sin ( 8 t ) .

The mass was released from the equilibrium position, so x ( 0 ) = 0 , and it had an initial upward velocity of 16 ft/sec, so x ( 0 ) = −16 . Applying these initial conditions to solve for c 1 and c 2 . gives

x ( t ) = −2 sin 8 t .

The period of this motion is 2 π 8 = π 4 sec.

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A 200-g mass stretches a spring 5 cm. Find the equation of motion of the mass if it is released from rest from a position 10 cm below the equilibrium position. What is the frequency of this motion?

x ( t ) = 0.1 cos ( 14 t ) (in meters); frequency is 14 2 π Hz.

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Writing the general solution in the form x ( t ) = c 1 cos ( ω t ) + c 2 sin ( ω t ) has some advantages. It is easy to see the link between the differential equation and the solution, and the period and frequency of motion are evident. This form of the function tells us very little about the amplitude of the motion, however. In some situations, we may prefer to write the solution in the form

x ( t ) = A sin ( ω t + ϕ ) .

Although the link to the differential equation is not as explicit in this case, the period and frequency of motion are still evident. Furthermore, the amplitude of the motion, A , is obvious in this form of the function. The constant ϕ is called a phase shift and has the effect of shifting the graph of the function to the left or right.

To convert the solution to this form, we want to find the values of A and ϕ such that

c 1 cos ( ω t ) + c 2 sin ( ω t ) = A sin ( ω t + ϕ ) .

We first apply the trigonometric identity

sin ( α + β ) = sin α cos β + cos α sin β

to get

c 1 cos ( ω t ) + c 2 sin ( ω t ) = A ( sin ( ω t ) cos ϕ + cos ( ω t ) sin ϕ ) = A sin ϕ ( cos ( ω t ) ) + A cos ϕ ( sin ( ω t ) ) .

Thus,

c 1 = A sin ϕ and c 2 = A cos ϕ .

If we square both of these equations and add them together, we get

c 1 2 + c 2 2 = A 2 sin 2 ϕ + A 2 cos 2 ϕ = A 2 ( sin 2 ϕ + cos 2 ϕ ) = A 2 .

Thus,

A = c 1 2 + c 2 2 .

Now, to find ϕ , go back to the equations for c 1 and c 2 , but this time, divide the first equation by the second equation to get

c 1 c 2 = A sin ϕ A cos ϕ = tan ϕ .

Then,

tan ϕ = c 1 c 2 .

We summarize this finding in the following theorem.

Practice Key Terms 3

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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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