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Let us describe this process in its general form.Suppose it is known that a force is acting on a point mass m that is free to movealong the x-axis, and that the force F = −kx is proportional33 to the displacement ofthe point from the origin. Suppose also that we know the initial position x0 = x(0) ofthe point mass and its initial velocity v0 = ẋ(0). Let us find the dependence x = x(t)of the position of the point on time.By Newton’s law, this problem can be rewritten in the following purely mathematical form: Solve the equationmẍ(t) = −kx(t)(5.156)under the initial conditions x0 = x(0), ẋ(0) = v0 .33 Inthe case of a spring, the coefficient k > 0 characterizing its stiffness is called the modulus.5.6 Examples of Differential Calculus in Natural Science299Let us rewrite Eq.
(5.156) asẍ(t) +kx(t) = 0m(5.157)and again try to make use of the exponential. Specifically, let us try to choose thenumber λ so that the function x(t) = eλt satisfies Eq. (5.157).Making the substitution x(t) = eλt in (5.157), we obtaink λt2e = 0,λ +morλ2 +k= 0,m(5.158)that is, λ1 = − − mk , λ2 = − mk . Since m > 0, we have the two imaginary numbers λ1 = −i mk , λ2 = i mk when k > 0. We had not reckoned on this possibility;however, let us continue our study. By Euler’s formula&&√kk−i k/mtt − i sint,e= cosmm&&√kkei k/mt = cost + i sint.mmSince differentiating with respect to the real variable t amounts to differentiatingthe real and imaginary parts ofthe functioneλt separately, Eq.
(5.157) must besatisfied by both functions cos mk t and sin mk t. And this is indeed the case, asone can easily verify directly. Thus the complex exponential function has enabledus to guess two solutions of Eq. (5.157), any linear combination of which&&kkx(t) = c1 cost + c2 sint,(5.159)mmis obviously also a solution of Eq. (5.157).We choose the coefficients c1 and c2 in (5.159) from the conditionx0 = x(0) = c1 ,&& &&&kkkk ksint + c2cost .v0 = ẋ(0) = −c1= c2mmmm t=0mThus the function&x(t) = x0 cosis the required solution.&&kmkt + v0sintmkm(5.160)3005Differential CalculusBy making standard transformations we can rewrite (5.160) in the form&&mk22x(t) = x0 + v0 sint +α ,(5.161)kmwhereα = arcsin x0.x02 + v02 mkThus, for k > 0 the point will make periodic oscillations with period T = 2π mk ,1kthat is, with frequency T1 = 2π,andamplitudex02 + v02 mk .
We state this bemcause it is clear from physical considerations that the solution (5.160) is unique.(See Problem 5 at the end of this section.)The motion described by (5.161) is called a simple harmonic oscillation, andEq. (5.157) the equation of a simple harmonic oscillator.Let us now turnto the case when k < 0in Eq. (5.158). Then the two functions eλ1 t = exp(− − mk t) and eλ2 t = exp( − mk t) are real-valued solutions ofEq. (5.157) and the functionx(t) = c1 eλ1 t + c2 eλ2 t(5.162)is also a solution.
We choose the constants c1 and c2 from the conditions!x0 = x(0) = c1 + c2 ,v0 = ẋ(0) = c1 λ1 + c2 λ2 .This system of linear equations always has a unique solution, since its determinant λ2 − λ1 is not 0.Since the numbers λ1 and λ2 are of opposite sign, it can be seen from (5.162)that for k < 0 the force F = −kx not only has no tendency to restore the point toits equilibrium position at x = 0, but in fact as time goes on, carries it an unlimiteddistance away from this position if x0 or v0 is nonzero.
That is, in this case x = 0 isa point of unstable equilibrium.In conclusion let us consider a very natural modification of Eq. (5.156), in whichthe usefulness of the exponential function and Euler’s formula connecting the basicelementary functions shows up even more clearly.Let us assume that the particle we are considering moves in a medium (the airor a liquid) whose resistance cannot be neglected. Suppose the resisting force isproportional to the velocity of the point. Then, instead of Eq. (5.156) we must writemẍ(t) = −α ẋ(t) − kx(t),which we rewrite asẍ(t) +αkẋ(t) + x(t) = 0.mm(5.163)5.6 Examples of Differential Calculus in Natural Science301If once again we seek a solution of the form x(t) = eλt , we arrive at the quadraticequationλ2 +√kαλ + = 0,mm2 −4mkαwhose roots are λ1,2 = − 2m± α 2m.The case when α 2 − 4mk > 0 leads to tworeal roots λ1 and λ2 , and the solutioncan be found in the form (5.162).We shall study in more detail the case in which we are more interested, whenα 2 − 4mk < 0.
Then both roots λ1 and λ2 are complex, but not purely imaginary:√4mk − α 2αλ1 = −−i,2m2m√4mk − α 2α+i.λ2 = −2m2mIn this case Euler s formula yieldsαλ1 te = exp −t (cos ωt − i sin(ωt),2mαλ2 te = exp −t (cos ωt + i sin ωt),2m√2α. Thus we find the two real-valued solutions exp(− 2m) cos ωtwhere ω = 4mk−α2mαand exp(− 2m ) sin ωt of Eq. (5.163), which would have been very difficult to guess.We then seek a solution of the original equation in the form of a linear combinationof these twoαt (c1 cos ωt + c2 sin ωt),(5.164)x(t) = exp −2mchoosing c1 and c2 so that the initial conditions x(0) = x0 and ẋ(0) = v0 are satisfied.The system of linear equations that results, as one can verify, always has a uniquesolution.
Thus, after transformations, we obtain the solution of the problem from(5.164) in the formαt sin(ωt + a),(5.165)x(t) = A exp −2mwhere A and a are constants determined by the initial conditions.αt), whenIt can be seen from this formula that, because of the factor exp(− 2mα > 0 and m > 0, the oscillations will be damped and the rate of damping ofα1. The frequency of the oscillations 2πω=the amplitude depends on the ratio m1kα 22πm − ( 2m ) will not vary over time.
The quantity ω also depends only on the3025Differential Calculusαratios mk and m, which, however, could have been foreseen from the form (5.163) ofthe original equation. When α = 0, we again return to undamped harmonic oscillations (5.161) and Eq. (5.157).5.6.7 Problems and Exercises1. Efficiency in rocket propulsion.a) Let Q be the chemical energy of a unit mass of rocket fuel and ω the outflow speed of the fuel.
Then 12 ω2 is the kinetic energy of a unit mass of fuel whenejected. The coefficient α in the equation 12 ω2 = αQ is the efficiency of the processes of burning and outflow of the fuel. For engines of solid fuel (smokeless powder) ω = 2 km/s and Q = 1000 kcal/kg, and for engines of liquid fuel (gasolinewith oxygen) ω = 3 km/s and Q = 2500 kcal/kg. Determine the efficiency α forthese cases.b) The efficiency of a rocket is defined as the ratio of its final kinetic energy2mR v2 to the chemical energy of the fuel burned mF Q. Using formula (5.139), obtaina formula for the efficiency of a rocket in terms of mR , mF , Q, and α (see part a)).c) Evaluate the efficiency of an automobile with a liquid-fuel jet engine, if theautomobile is accelerated to the usual city speed limit of 60 km/h.d) Evaluate the efficiency of a liquid-fuel rocket carrying a satellite into loworbit around the earth.e) Determine the final speed for which rocket propulsion using liquid fuel ismaximally efficient.f) Which ratio of masses mF /mR yields the highest possible efficiency for anykind of fuel?2.
The barometric formula.a) Using the data from Sect. 5.6.2, obtain a formula for a correction term to takeaccount of the dependence of pressure on the temperature of the air column, if thetemperature is subject to variation (for example, seasonal) within the range ±40 °C.b) Use formula (5.144) to determine the dependence of pressure on elevation attemperatures of −40 °C, 0 °C, and 40 °C, and compare these results with the resultsgiven by your approximate formula from part a).c) The change in temperature of the atmosphere at an altitude of 10–11 km iswell described by the following empirical formula: T (h) = T0 − αh, where T0 is thetemperature at sea level (at h = 0 m), the coefficient α = 6.5 × 10−3 K/m, and h isthe height in meters. Deduce under these conditions the formula for the dependenceof the pressure on the height.
(T0 it is often given the value 288 K, corresponding to15 °C.)d) Find the pressure in a mine shaft at depths of 1 km, 3 km and 9 km usingformula (5.144) and the formula that you obtained in c).5.6 Examples of Differential Calculus in Natural Science303e) Independently of altitude, air consists of approximately 1/5 oxygen. The partial pressure of oxygen is also approximately 1/5 of the air pressure. A certainspecies of fish can live under a partial pressure of oxygen not less than 0.15 atmospheres. Should one expect to find this species in a river at sea level? Could it befound in a river emptying into Lake Titicaca at an elevation of 3.81 km?3. Radioactive decay.a) By measuring the amount of a radioactive substance and its decay products inore samples of the Earth, assuming that no decay products were originally present,one can estimate the age of the Earth (at least from the time when the substanceappeared).
Suppose that in a rock there are m grams of a radioactive substance andr grams of its decay product. Knowing the half-life T of the substance, find the timeelapsed since the decay began and the amount of radioactive substance in a sampleof the same volume at the initial time.b) Atoms of radium in an ore constitute approximately 10−12 of the total numberof atoms. What was the radium content 105 , 106 , and 5 × 109 years ago? (The ageof the Earth is estimated at 5 × 109 years.)c) In the diagnosis of kidney diseases one often measures the ability of the kidneys to remove from the blood various substances deliberately introduced into thebody, for example creatin (the “clearance test”). An example of an opposite processof the same type is the restoration of the concentration of hemoglobin in the bloodof a donor or of a patient who has suddenly lost a large amount of blood.
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