THE LENGTHENING PENDULUM

18. THE LENGTHENING PENDULUM

As an example of the use of Bessel functions we consider the following problem. Suppose that a simple pendulum (see Chapter 11, Section 8) has the length l of its string increased at a steady rate (for example, a weight swaying as it is lowered by a crane). (This problem was considered as early as 1707; see L. LeCornu, Acta Mathematica

19 (1895), 201–249. Also see Relton, and Problem 8.) Find the equation of motion and the solution for small oscillations. From Chapter 11, Section 8, we have the equation of motion

Let the length of the string at time t be (18.2)

l=l 0 + vt,

and change from t to l as the independent variable. For small oscillations, we may replace sin θ by θ. Then (18.1) becomes (Problem 1):

(This equation could also describe the damped vibration of a variable mass, or an RLC circuit with variable L.)

We solve (18.3) by comparing it with the standard equation (16.1) to get (Prob- lem 2)

where b = 2g 1/2 /v. To simplify the notation, let (18.5)

θ=l −1/2 Z 1 (bl 1/2 )

u=bl 1/2 = (2g 1/2 /v)l 1/2 .

The general solution of (18.3) is then (18.6)

θ = Au −1 J 1 (u) + Bu −1 N 1 (u).

We can find dθ/du from (18.6) using (15.2):

dθ

= −[Au −1 J 2 (u) + Bu −1 N 2 (u)].

du

Section 18 The Lengthening Pendulum 599

The constants A and B must be found from the starting conditions just as they are for the ordinary simple pendulum with constant l. For example, in the ordinary case, if θ = θ 0 and ˙θ = 0 at t = 0, then the general solution θ = A cos ωt + B sin ωt becomes just θ = θ 0 cos ωt. For the lengthening pendulum, let’s take the same simple initial conditions, namely θ = θ 0 and ˙θ = 0 at t = 0. For these initial conditions, we find (after some calculations—see Problems 3 to 6)

The solution has a particularly simple form if we adjust the constants v and l 0 so that

is a zero of J 2 (u). Then B = 0 and the second term of (18.6) is zero, so we have (18.10)

u 0 = 2(gl 0 ) 1/2 /v

θ = Au −1 J 1 (u) = Cl −1/2 J 1 (b l 1/2 ),

where (Problem 7)

For this simple case, ˙θ is a multiple of J 2 (u) (Problem 8); thus θ = 0 corresponds to zeros of J 1 (u) and ˙θ = 0 corresponds to zeros of J 2 (u). A “quarter” period corresponds to the time from θ = 0 to ˙θ = 0, or ˙θ = 0 to θ = 0. These quarter periods can be found from the zeros of J 1 (u) and J 2 (u) (Problem 8).

PROBLEMS, SECTION 18

1. Verify equation (18.3). Hint: From equation (18.2), dl = v dt, so

2. Solve equation (18.3) to get equation (18.4). 3. Prove

J p (x)J ′ −p (x) − J −p (x)J p ′ (x) = −

sin pπ

πx

as follows: Write Bessel’s equation (12.1) with y = J p and with y = J −p ; multiply the J p equation by J −p and the J −p equation by J p and subtract to get

[x(J p J ′

−p −J −p J p ′ )] = 0.

dx

Then J p J −p ′ −J −p J p ′ = c/x. To find c, use equation (12.9) for each of the four functions and pick out the 1/x terms in the products. Then use equation (5.4) of Chapter 11.

4. Using equation (13.3) and Problem 3, show that

p (x)J ′ (x) 2 J p (x)N p ′ (x) − J p ′ (x)N p (x) = p

J ′ (x)J

−p (x) − J

−p

sin pπ

πx

600 Series Solutions of Differential Equations Chapter 12

5. Use the recursion relations of Section 15 (for N ’s as well as for J’s) and Problem 4 to show that

J n (x)N n+1 (x) − J n+1 (x)N n (x) = −

. πx

Hint: Do it first for n = 0; then use the result in proving the n = 1 case, and so on. 6. For the initial conditions θ = θ 0 , ˙θ = 0, show that the constants A and B in

equations (18.6) and (18.7) are as given in (18.8). Hints: Show that dθ/du = 0 if ˙θ = 0. In equations (18.6) and (18.7), set θ = θ 0 and dθ/du = 0 when u = u 0 and solve for A and B. Then use the formula in Problem 5 to simplify your results to get equation (18.8).

7. Verify the values of b and C given in equation (18.11). Note that C can be found in two ways: (1) in equation (18.10), u = bl 1/2 , so Au −1 = (A/b)l −1/2 , C = A/b. Use

Problem 5 to simplify this. (2) Set θ = θ 0 ,u=u 0 ,l=l 0 in equation (18.10) and solve for C.

either from equations (18.10) and (15.2) or from equation (18.7) with B = 0. Thus show that θ = 0 when J 1 (u) = 0 and ˙θ = 0 when J 2 (u) = 0. Show that the successive (variable) quarter periods of the lengthening pendulum are (v/4g)(r 2 2 2 −r 1 ) or (v/4g)(r 2 1 2 −r 2 ), where r 1 and r 2 are successive zeros of J 1 and J 2 . Use a computer or tables to find the needed zeros and calculate several quarter periods (as multiples of v/(4g)). Observe that an inward swing takes longer than either the preceding or the following outward swing. [This result is proved by Ll. G. Chambers, Proceedings of the Edinburgh Mathematical Society (2) 12, 17–18 (1960).]

9. Consider the “shortening pendulum” problem. Follow the method in the text but with l = l 0 − vt. Does the θ amplitude of the vibration increase or decrease as the pendulum shortens? Restate the result of Problem 8 about quarter periods for this case.

10. The differential equation for transverse vibrations of a string whose density increases linearly from one end to the other is y ′′ + (Ax + B)y = 0, where A and B are constants. Find the general solution of this equation in terms of Bessel functions. Hint: Make the change of variable Ax + B = Au.

11. A straight wire clamped vertically at its lower end stands vertically if it is short, but bends under its own weight if it is long. It can be shown that the greatest length for vertical equilibrium is l, where kl 3/2 is the first zero of J −1/3 and

r = radius of the wire, ρ = linear density, g = acceleration of gravity, Y = Young’s modulus. Find l for a steel wire of radius 1 mm; for a lead wire of the same radius.

Section 19 Orthogonality of Bessel Functions 601