Physics 181

Assignment 6

1. A particle of mass m slides without friction on a track that is part of a circle of radius R as shown. We are interested in the small oscillations about the bottom of the track.

a Taking U = 0 at the bottom of the track. write U as a function of the angle θ .

The height of the particle above the bottom of the track is y = R(1 − cosθ) so we have

U(θ) = mgR(1 − cosθ) .

b. Let s be the distance the particle moves along the track from the bottom. What is s in terms of R and θ ?

It is the arc of the circle, so s = Rθ .

c. Use the small angle approximation for the cosine, cosθ ≈ 1 −θ 2 /2 to show that for small oscillations the motions is simple harmonic, i.e., the potential energy is approximately of the form U(s) = 1

2 ks2 .

We have U = mgR[1 − cos(s/R)] ≈ mgR[(s/R)2 /2] = 12 m(g/R) ⋅ s2 .

d. What is the angular frequency ω of these small oscillations?

The standard form is U = 12 mω 2 ⋅ s2 , so we find ω = g/R . It is like a pendulum of

length R.

e. Is the actual frequency larger or smaller than this? [What is the next term in the Taylor series for cosθ ? Does it make the restoring force −dU / ds stronger or weaker?]

θ R

1

Including the next term we have cosθ ≈ 1 − 12θ

2 + 124θ

4 . This leads to

U(s) = 12 mgR[(s/R)2 − 1

12 (s/R)4] . We find F(s) = −dU /ds = −mg[(s/R) − 13 (s/R)3] . The

restoring force is weaker, so the frequency is smaller. It’s like having a weaker spring.

2. It is possible to shape the track such that the potential energy as a function of distance s along the track is exactly of the form U(s) = 1

2 ks2 , which will make the

oscillations purely simple harmonic.

a. Use a cartesian system with origin at the bottom of the track. Let the curve of the track be that of a cycloid, defined parametrically by the equations

x = R(α + sinα )y = R(1 − cosα )

where R is a constant and −π ≤ α ≤ π is a variable angle. Use the fact that an infinitesimal distance ds along the track is given by ds2 = dx2 + dy2 .

Express ds/dα as a function α . [Use the identity 1 + cosα = 2cos2(α/2) .]

We have dx = R(1 + cosα )dα and dy = Rsinα dα . Squaring and adding we find

ds2 = 2R2(1 + cosα )dα 2 = 4R2 cos2(α/2)dα 2 , or ds/dα = 2Rcos(α/2) .

b. Integrate this to find s as a function of α .

Integrate: s = (ds/dα )dα

0

α∫ = 2R cos(α/2)dα

0

α∫ = 4Rsin(α/2) .

c. Use your result to express y in terms of s. [Use the identity

2sin2(α/2) = 1 − cosα .]

We have s2 = 16R2 sin2(α/2) = 8R2(1 − cosα ) = 8R ⋅ y . So y = s2 /8R .

d. Write the potential energy of the system in terms of s.

Choosing the bottom of the track to have zero potential energy, U = mgy = mgs2 /8R .

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e. Find the angular frequency ω of the oscillations.

The standard form is U = 12 mω 2s2 , so we see that

ω = 1

2gR

.

3. Consider the driven oscillator described in the notes on Differential Equations, page 10. Use the equations given there.

a. Carry out the integral necessary to derive the equation Pin( )av =

F02

2bsin2δ .

[The average over a cycle of a function f (t) is defined by fav =

1T

f (t)dt0

T∫ .

For our case T = 2π /ω , the period of the oscillation.]

The expression given in the notes is Pin(t) = F0 ⋅ωA ⋅[sinωtcosωtcosδ + sinωtsinωtsinδ] .

The average over a cycle of this is

ω2π

⋅F0ωA ⋅ cosδ ⋅ sinωt ⋅cosωt dt0

2π/ω∫ + sinδ ⋅ sin2ωt dt

0

2π/ω∫⎡

⎣⎢⎤⎦⎥

. Let θ =ωt . Then the first

integral becomes 1ω

sinθ ⋅cosθ dθ0

2π∫ , which gives zero. The second integral is

1ω

sin2θ dθ0

2π∫ , which gives π /ω . So we have (Pin )av = 1

2 F0ωAsinδ . But, as the

formulas in the notes show, ωA = (F0 /b)sinδ , so we obtain the desired formula.

b. Let F0 = 1 , m = 1 , ω0 = 1 . Consider two cases: (1) b = 0.5 ; (2) b = 2 . Use a

graphing calculator and sketch the curve of (Pin )av vs. ω for both cases on

the same plot.

Web sites have graphing calculators that can be used free. One is found at http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html

4. A particle of mass m moves in the x-y plane with potential energy given by

U(x, y) = 12 kx2 + 2ky2 .

The particle is originally at rest at the point where x = y = A .

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a. Write the Lagrange equations for the variables x and y.

We have L = 1

2 m( x2 + y2 ) − 12 k(x2 + 4y2 ) . This leads to the equations mx + kx = 0 and

my + 4ky = 0 .

b. Solve for x(t) and y(t) . [Use the initial conditions and the notation

ω = k/m .]

The general solutions are x = Ax cos(ωt + δx ) and y = Ay cos(2ωt + δy ) . The conditions

x(0) = 0 and y(0) = 0 require both δ ’s to be zero. Then x(0) = y(0) = A gives the final

results x = Acos(ωt) and y = Acos(2ωt) .

c. Derive the equation of the trajectory, i.e., y = f (x) . [Use the identity

cos2α = 2cos2α − 1 .]

Using the hint we see that y = A[2cos2(ωt) − 1] = 2

Ax2 − A . This is the equation of a

parabola.

d. Sketch the trajectory for the values m = k = A = 1 . Describe the motion as a function of time.

Shown below is the curve y = 2x2 − 1 . The particle starts at the upper right end, runs

around the curve to the opposite end, and back again.

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5. Two identical blocks of mass m and two identical springs of stiffness k are arranged as shown. The left spring is attached to a fixed wall, and the blocks can slide on the floor without friction. Let the displacement (to the right) of the left block be x1 and that of the right block x2 .

a. Write the potential energy of the system, and the Lagrange equations for the x’s. Use the notation k = mω0

2 .

We have U = 12 mω0

2[x12 + (x2 − x1)2] = 1

2 mω02[2x1

2 + x22 − 2x1x2] . The Lagrange equations

give x1 +ω0

2(2x1 − x2 ) = 0 , x2 +ω0

2(x2 − x1) = 0 .

5

x

y

b. Assume solutions of the form x1 = Aeiωt , x2 = Beiωt . Find the angular

frequencies ω1 and ω2 of the two normal modes. [Choose ω1 >ω2 .]

We have A(2ω02 −ω 2 ) − Bω0

2 = 0 and −Aω02 + B(ω0

2 −ω 2 ) = 0 . Setting the determinant of

the coefficients to zero we find ω4 − 3ω0

2ω 2 +ω04 = 0 . The roots are

ω1

2 = 12 3 + 5( )ω0

2

and ω2

2 = 12 3 − 5( )ω0

2 .

c. For ω =ω1 , find the ratio B / A . Repeat for ω =ω2 .

Substituting ω1 into B(ω02 −ω 2 ) = Aω0

2 we find

BA

= −2

5 + 1= −

5 − 12

. Using ω2 we

find

BA

=2

5 − 1=

5 + 12

.

d. The normal coordinates (up to a multiplicative constant) have the form

q = x1 + cx2 . Evaluate the c’s for ω1 and ω2 , thereby finding the two

normal coordinates q1 and q2 .

Take A = 1 in the above calculations. Then for ω1 we find q1 = x1 −

5 − 12

x2 . For ω2 we

find q2 = x1 +

5 + 12

x2 . [To get the standard normal coordinates, one must choose the

A’s so that q12 + q2

2 = x12 + x2

2 .]

e. Suppose at t = 0 we have x1 = A . For what initial value of x2 will the

system oscillate in the normal mode corresponding to ω1 ? Repeat for ω2 .

If x2 = −

5 + 12

A , then q2 = 0 and the motion is entirely the first normal mode. If

x2 = +

5 − 12

A , then q1 = 0 and the motion is entirely the second normal mode.

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