Query 18

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Good question, and fortunately the answer is pretty simple.

You also know that the velocity at the initial instant is zero. So y ' (0) = 0, which will lead you to the conclusion that c2 must be 0.

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You don't need to relate omega to 2 pi in order to solve this problem, but it is still important to understand the relationship.

Your solution consists of sines and cosines of 326 t.

The sine and cosine both go through their periods when then arguments change by 2 pi.

So when 326 t changes by 2 pi, you complete a full cycle.

326 t changes by 2 pi when t changes by 2 pi / 326, so the time required to complete a cycle is `dt = 2 pi / 326.

This is called the period of the oscillation.

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All that being said, note that the solution to

y1^2 = -326

is

y1 = sqrt(326) * i = 18 i, approximately.

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omega isn't the period.

Per the note inserted in the preceding problem, the period is 2 pi / omega, so omega is 2 pi / period.

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To understand this you need a little better understanding of the graphs of sine and cosine functions. But you're on the right track.

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The period of this function is 2, so omega = 2 pi / 2 = pi.

The amplitude is 3 so the function is of the form y = 3 cos(pi t - delta).

The graph appears to be shifted about 1/4 unit to the right of the graph of y = 3 cos(pi t); replacing t by t - 1/4 will accomplish the shift to the right. Our function is therefore

y = 3 cos( pi ( t - 1/4) ) ) = 3 cos( pi t - pi/4).

Thus delta = pi/4, omega = pi and R = 3.

**** not enough info: The mass is m, and omega = sqrt(k / m) so k = m omega^2 = 4 m. ****

The initial condition on y is y(0) = 2.

y ' (t) = (3 cos( pi t - pi/4)) ' = -2 pi sin( pi t - pi/4) so

y ' (0) = -3 pi sin( -pi/4) = 3 pi sqrt(2) / 2 = 6.5, approximately.

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This solution doesn't follow from your equation. You get c2 = .04 / 326, not 326 / .04.

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Good, except for errors in detail.

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With details corrected I believe the correct solution is

y(t) = -30 cos( 18 t) - 2.3 sin(18 t).

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The equation would be

y '' + gamma y ' + k/m y = 0.

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The system is critically damped. So the discriminant

sqrt( gamma^2 - 4 k / m)

is zero, implying that gamma = 2 sqrt(k/m).

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See also my notes above. You're doing OK and asking the right questions.

Here's a detailed solution:

If y(t) is the position of the object, the damping force is of form F_damp = - gamma y ' and the spring force is F_spring = - k y so that

F_net = -gamma y ' - k y

and

m y '' + gamma y ' + k y = 0

or

y '' + gamma / m * y ' + k / m * y = 0.

(The unit of mass in the British system is the slug; the weight of 1 slug is 1 slug * 32 ft / s^2 = 32 slug * ft / s^2 = 32 pounds, so our weight has mass 1 slug.

6 inches is 1/2 foot.

y is in units of ft , y ' in units of ft / s^2 and y '' in units of ft / s^2

k * y is in units of lb, so k is in units of lbs / ft, k / m is in units of (lbs / ft) / slugs, which is reduces to s^2

gamma * y ' is in units of pounds, so gamma is in units of lbs / (ft / s) = kg s^-1, so gamma / m is in units of s^-1.

Our conditions are

y(0) = 0

y ' (0) = -4 ft / sec

y(t) = -1/2 ft when | y(t) | is maximized).

The characteristic equation is

r^2 + gamma / m * r + k / m = 0

with solutions

r = (-gamma / m +- sqrt( gamma^2 / m^2 - 4 k / m) ) / 2.

The system is critically damped so the discriminant is 0:

gamma^2 / m^2 - 4 k / m = 0

with the result that

gamma = +-2 sqrt(k * m).

gamma is positive, so we discard the negative solution.

It follows that

r = -gamma / (2 m) = -2 sqrt(k * m) / (2 m) = -sqrt ( k / m )

and our fundamental set is

{e^(-sqrt( k / m) * t, t e^(-sqrt(k / m) * t) }.

Note that we generally use omega is these solutions, where omega = sqrt(k / m). However omega has a strong connotation with angular frequency, and our solutions in the critically damped (and overdamped) cases are not oscillatory so, to avoid possible confusion, we'll leave the notation as sqrt(k / m).

Our general solution is

y(t) = A e^(-sqrt(k / m) * t) + B t e^(-sqrt(k / m) * t).

Initially y = 0 s0

y(0) = 0

giving us

A e^(-sqrt(k / m) * 0) + B * 0 * e^(-sqrt(k / m) * 0) = 0

with the result that

A = 0

giving us

y(t) = B t e^(-sqrt(k / m) * t).

We also note for future reference that

y ' (t) = B e^(-sqrt(k / m) * t) - B t sqrt(k/m) * e^(-sqrt(k / m) * t).

The initial velocity is 4 ft/sec downward so

y ' (0) = - 4 ft / sec

giving us

B e^(-sqrt(k / m) * 0) = -4

so that

B = -4.

Our function is thus

y = -4 t e^(-sqrt(k / m) * t)

and

y ' = -4 e^(-sqrt(k / m) * t) + 4 t sqrt(k/m) * e^(-sqrt(k / m) * t).

(note on units: it should be clear that B is in units of ft / sec; k/m is in units of s^-2 so sqrt(k/m) is in units of s^-1, which is appropriate since t is in units of s)

We are given the weight of the object, and as noted earlier its mass is easily found; in this case the mass is 1 slug.

Our equation is y = -4 t e^(-sqrt(k / m) * t); the only remaining unknown quantity is k.

To evaluate k we use the last bit of given information:

y = -1/2 ft is the position at which distance from equilibrium is maximized.

This occurs when the mass comes to rest, so we have

y(t) = 1/2 when y ' (t) = 0.

y ' (t) = -4 e^(-sqrt(k / m) * t) + 4 sqrt(k / m) t e^(-sqrt(k / m) * t)

y ' (t) = 0 when

y ' = -4 e^(-sqrt(k / m) * t) + 4 t sqrt(k/m) * e^(-sqrt(k / m) * t)

so we solve

-4 e^(-sqrt(k / m) * t) + 4 t sqrt(k/m) * e^(-sqrt(k / m) * t) = 0

for t, obtaining

(4 - 4 t sqrt(k/m) ) e^(-sqrt(k / m) * t) = 0

so that

t = 1 / sqrt(k / m).

So the maximum displacement from equilibrium occurs when t = 1 / (sqrt(k/m)), and at that instant y = -1/2 foot. Writing this as an equation

y(1 / sqrt(k/m)) = -1/2 so

-4 (1 / sqrt(k/m)) e^(-sqrt(k/m) * (1 / (sqrt(k/m) ) ) = -1/2

-4 e^-1 = -1/2 sqrt(k/m)

sqrt(k/m) = 8 / e

(With units the equation -4 e^-1 = -1/2 sqrt(k/m) reads -4 ft / s * e^-1 = -1/2 ft * sqrt(k/m), so that sqrt(k/m) is in units of (ft / s) / ft = s^-1. This is consistent with k in units of lb / ft and mass in slugs (see the earlier note summarizing units)).

k = 8 m / e

(We note that k is therefore in units of mass / time, appropriate units for a force constant).

As seen before gamma = sqrt( 2 k m) so

gamma = sqrt( 2 * (8 m / e) * m ) * m) = 4 m sqrt(1/e).

(the units of k * m are lbs / ft * slugs = kg^2 s^-2, so the units of sqrt(k m) and therefore gamma are kg s^-1, consistent with the units as noted early in the solution).

We calculate our numerical values of k and gamma:

m = 1 (mass is 1 slug) so

k = 8 / e = 2.9 kg / s^2 = 2.9 lb / ft

gamma = 4 m sqrt(1/e) kg s^-1 = 1.45 kg s^-1, or 1.45 lb / (ft / s).

For every foot of stretch the spring exerts 2.9 pounds, and for every ft / sec of velocity the drag force is .30 lb.

Checking against common sense:

The initial drag force is 1.45 lb / (ft / s) * 4 ft /s = 6 lb, approx.. This would result in an initial acceleration of about 6 ft / s^2 on our 1-slug mass.

The maximum spring force is 2.9 lb * 1/2 ft = 1.45 lb.

The time required to reach max displacement is 1 / sqrt(k / m) = .6 second

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As in the preceding the characteristic equation has solutions

r = (-gamma / m +- sqrt( gamma^2 / m^2 - 4 k / m) ) / 2 = -gamma / (2 m) +- gamma / (2 m) * sqrt( 1 - 4 k m / (gamma^2) ) = -gamma / (2 m) * (1 +-sqrt( 1 - 4 k m / gamma^2) ).

The discriminant is:

zero when gamma = 2 sqrt(k * m),

negative when gamma < 2 sqrt( k * m )and

positive when gamma > 2 sqrt( k * m ) ,

corresponding respectively to the critically damped, overdamped and underdamped cases.

The solution sets are respectively

{e^(-gamma / (2 m) * t), t e^(-gamma / (2 m) * t) }

{e^(-gamma / (2 m) * (1 +sqrt( 1 - 4 k m / gamma^2)) * t), e^(-gamma / (2 m) * (1 -sqrt( 1 - 4 k m / gamma^2)) * t)}

{e^(-gamma / (2 m) * (1 + i sqrt( 4 k m / gamma^2 - 1)) * t), e^(-gamma / (2 m) * (1 - i sqrt( 4 k m / gamma^2 - 1)) * t), which leads to

{e^(-gamma / (2 m) * t) cos(sqrt( 4 k m / gamma^2 - 1)) * t), e^(-gamma / (2 m) * t) sin( sqrt( 4 k m / gamma^2 - 1)) * t)}.

The general solutions can be written as follows (in each case the common factor e^(-gamma / (2 m) * t) has been factored out:

Underdamped case:

e^(-gamma / (2 m) * t) * ( A cos(sqrt( 4 k m / gamma^2 - 1)) * t) + B sin( sqrt( 4 k m / gamma^2 - 1)) * t )

Critically damped case (here we have used C and D as our constants to distinguish them from the constants A and B of the case with which this one is being compared):

e^(-gamma / (2 m) * t) * ( C + D t)

Overdamped case:

e^(-gamma / (2 m) ) * ( A e^(1 +sqrt( 1 - 4 k m / gamma^2)) * t) + B e^((1 -sqrt( 1 - 4 k m / gamma^2)) * t )

Observe that each solution contains the factor e^(-gamma / (2 m) ). We therefore need to show that

as gamma approaches 2 sqrt( k * m ) from above, A cos(sqrt( 4 k m / gamma^2 - 1)) * t) + B sin( sqrt( 4 k m / gamma^2 - 1)) * t approaches the form C + D * t

as gamma approaches 2 sqrt( k * m ) from below, A e^(1 +sqrt( 1 - 4 k m / gamma^2)) * t) + B e^((1 -sqrt( 1 - 4 k m / gamma^2)) * t approaches the form C + D * t

Now, note that

as gamma approaches 2 sqrt( k * m ) from above, the quantity 4 k m / gamma^2 approaches 1 from below so that 1 - 4 k m / gamma^2 approaches zero from above

as gamma approaches 2 sqrt( k * m ) from below, the quantity 4 k m / gamma^2 approaches 1 from above so that 4 k m / gamma^2 - 1 approaches zero from above

We make use of the following facts, easily proven by considering the first two terms of the Taylor series of sin(x), sqrt( x) and e^x:

If epsilon is a small number, then e^(1 +- epsilon) is close to 1 +- epsilon.

If epsilon is a small number, then sin(x) is close to x.

Consider first the overdamped case.

A e^(1 +sqrt( 1 - 4 k m / gamma^2)) * t) + B e^((1 -sqrt( 1 - 4 k m / gamma^2)) * t )

The quantity sqrt( 1 - 4 k m / gamma^2)) * t does approach zero from above, meaning that this quantity approaches zero through positive values.

Letting epsilon = sqrt( 1 - 4 k m / gamma^2)) * t, then our first term is approximately equal to

A * ( 1 + sqrt( 1 - 4 k m / gamma^2)) * t)

and our second term is

B * (1 - sqrt( 1 - 4 k m / gamma^2)) * t )

so that our solution

A e^(1 +sqrt( 1 - 4 k m / gamma^2)) * t) + B e^((1 -sqrt( 1 - 4 k m / gamma^2)) * t )

is approximately equal to

A + B + ( A - B ) sqrt( 1 - 4 k m / gamma^2)) * t

Now let epsilon = sqrt( 1 - 4 k m / gamma^2)). Since sqrt( 1 - epsilon) = 1 - epsilon/2, we have

sqrt(1 - 4 k m / gamma^2) = 1 - 2 k m / gamma^2, approximately. So

A + B + ( A - B ) sqrt( 1 - 4 k m / gamma^2)) * t = (A + B) + (A - B) ( 1 - 2 k m / gamma^2) * t

and as gamma approaches 2 sqrt(k * m) this approaches

A + B + (A - B) * 1/2 * t

which is of the form C + D t for C = A + B, and D = 1/2 ( A - B ).

Consider next the underdamped case.

A cos(sqrt( 4 k m / gamma^2 - 1)) * t) + B sin( sqrt( 4 k m / gamma^2 - 1)) * t.

As gamma approaches 2 sqrt(k m) the term sqrt( 4 k m / gamma^2 - 1)) approaches zero from above, so that

cos(sqrt( 4 k m / gamma^2 - 1))* t) approaches 1

and

sin(sqrt( 4 k m / gamma^2 - 1)) * t) approaches sqrt( 4 k m / gamma^2 - 1)), which in turn approaches 2 k m / gamma^2.

Thus

A cos(sqrt( 4 k m / gamma^2 - 1)) * t) + B sin( sqrt( 4 k m / gamma^2 - 1)) * t

approaches

A + 2 k m / gamma^2 * B * t

which is of the form

C + D t

for

C = A

and

D = 2 k m / gamma^2 * B.

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&#Good responses. See my notes and let me know if you have questions. &#