Query 18

Good, but the mass is pulled down 70 mm from its unloaded position, which is only 40 mm below its loaded equilibrium position.

@& For comparison:

A 10 kg mass weighs 98 Newtons. Thus a 98 N change in force will change the length of the spring by 30 mm. Assuming the force to be a linear function of the spring we find that the force constant is

k = 98 N / (30 mm) = 3.3 N / mm, approx., or about 3300 N / m.

The new equilibrium position of the system will be 30 mm below the unloaded position. When pulled down 70 mm the position of the system will therefore be -40 mm.

Thus, if y is the position relative to the unloaded position, y(0) = -40 mm and since the spring is released from rest, y ' (0) = 0.

The net force on the system is

F = -k y so that

m y '' = - k y and

y '' = -k/m * y.

The general solution to this equation is

y = A cos(omega t) + B sin(omega t)

with omega = sqrt(k / m) = sqrt(3300 N/m / (10 kg) ) = 18 rad / sec, approx..

We get

y ' = -omega A sin(omega t) + omega B cos(omega t)

The initial conditions give us

y(0) = A = -40 mm

and

y ' (0) = omega B = 0

so that

A = -40 mm, B = 0 and our solution is

y = -40 mm * cos(omega * t).

*@

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@& The cosine is -1 at pi, not 2 pi.*@

@& Good, but the pi vs. 2 pi creates a discrepancy.

See if you agree with the following:

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(2 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 - pi/4 will accomplish the shift to the right. Our function is therefore

y = 3 cos( 2 ( t - pi/4) ) ) = 3 cos( 2 t - pi/2).

Thus delta = pi/2, omega = 2 and R = 3.

*@

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Considering that the loaded equilibrium position is the 40 mm position see if you agree with the following:

Recall that the equilibrium position for the oscillator is 40 mm below its unloaded position.

The 70 mm position is 30 mm below equilibrium, so that initially y = -30 mm, giving us the initial condition

y(0) = -30 mm.

The initial velocity is 40 mm / sec downward, so that

y ' (0) = -40 mm/sec.

As before the equation of motion is

y '' = -k/m * y.

and its general solution is

y(t) = A cos(omega t) + B sin(omega t)

with omega = sqrt(k / m).

y(0) = -30 mm yields A = -30 mm.

y ' (t) = -omega A sin(omega t) + omega B cos(omega t)

so that

y ' (0) = omega * B.

Recall that omega = 18 rad / sec, approx., so

y ' (0) = 18 rad / sec * B = -40 mm / sec

so that

B = -40 mm/s / (18 rad /s) = -2.3 mm.

The equation of motion for the mass is therefore

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

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The 6-inch distance would be expressed as 1/2 foot, to make its units consistent with the other units in the problem. This would affect your value of lambda.

Compare with the following, which follows a slightly different (but equivalent) track:

@& 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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This one isn't bad if you consider Taylor series of exponentials, sines, and square roots near 1.

@&

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.

*@

@& You're doing well. Check my notes for a little more on some of these solutions.*@