MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

8.01

Problem Solving Session 10 Simple Harmonic Oscillator Solutions W13D3-0 Group Problem Gravitational Simple Harmonic Oscillator

Two identical point-like objects each of mass m1 are fixed in place and separated by a distance 2d . A third object of mass m2 is free to move and lies on the perpendicular bisector of the line connecting the two objects at a distance y(t) above the midpoint of that line. The objects interact gravitationally. The force is attractive and has magnitude

!Fgrav =

Gm1m2

r2.

where r is the distance between the objects.

a) Determine an expression for the y –component of the gravitational force acting on the object in terms of m1 , m2 , G , ( )y t , and d as needed using the

fact that cos! = y / (d 2 + y2 )1/ 2 .

b) Find an equation that the acceleration d2 y / dt2 of the object with mass m2

satisfies in terms of the quantities m1 , m2 , G , ( )y t , and d as needed.

c) Use the approximation that y(t) << d in your equation in part a) to find an equation for d

2 y / dt2 in terms of the quantities m1 , m2 , G , ( )y t , and d as needed.

d) Assume that at 0t = , the object of mass m2 is at the equilibrium position but

has an initial upward speed v0 . Write down an expression for the subsequent motion ( )y t , in terms of m1 , m2 , G , v0 , and d as needed. Assume that

y(t) << d .

Solution: The free body force diagram on the object of mass m2 is shown in the figure below

The y –component of force acting on the object is

Fy = !

Gm1m2

(d 2 + y2 )2cos" .

From the diagram cos! =

y(d 2 + y2 )1/ 2

so the y –component of force is

Fy = !

2Gm1m2 y(d 2 + y2 )3/ 2

.

Therefore Newton’s Second Law yields the equation of motion

m2

d 2 ydt2 = !

2Gm1m2 y(d 2 + y2 )3/ 2

When y(t) << d the equation of motion becomes

d 2 ydt2 ! !

2Gm1

d 3 y .

This is a simple harmonic oscillation equation of motion with solution

y(t) = y0 cos(!0t) +

vy ,0

!0

sin(!0t)

where

!0 =

2Gm1

d 3 .

The initial conditions are y0 = 0 and

vy ,0 = v0 > 0 . Therefore

y(t) =

v0

!0

sin(!0t) ,

vy (t) = v0 cos(!0t) .

IC_W13D3-3 Group Problem Kater’s Pendulum Solutions A pendulum consists of a rod and two knife-edges separated from the center of mass l1 and l2 respectively (figure below left). The moment of inertia of the rod about the center

of mass is I0 = mkg

2 where kg is a constant called the radius of gyration. When the

pendulum is pivoted about the upper knife-edge (figure below center), the period for small oscillations is T1 . When the pendulum is turned upside down and pivoted about the other edge (figure below right), and the period for small oscillations is T2 . The distances

l1 and l2 are adjusted until T1 = T2 ! T . Express your answers to the following questions in terms of l1 , l2 , and T as needed.

a) What is the radius of gyration kg ?

b) Using your results from part (a), show that the gravitational acceleration g

can be determined by measuring the length between the knife-edges l1 + l2 and the period T .

Solution: We can treat each case as a physical pendulum. The torque equation of motion when suspended from the upper edge is

!l1mg sin" = I1

d 2"dt2 .

The moment on inertia about the upper knife-edge is

I1 = I0 + ml1

2 = m(kg2 + l1

2 ) . Using the small angle approximation sin! ! ! . The torque equation becomes

!l1mg" = m(kg

2 + l12 )

d 2"dt2 .

Then the equation of motion for the pendulum is a simple harmonic oscillator equation of motion

d 2!dt2 +

l1g(kg

2 + l12 )! = 0 .

The period of oscillation is therefore

T1 = 2! (kg

2 + l12 ) / l1g .

When the pendulum is turned upside down and suspended from the other knife-edge and similar argument yields for the period

T2 = 2! (kg

2 + l22 ) / l2g .

Because distances l1 and l2 have been adjusted until T1 = T2 ! T , we can solve for

kg by squaring the periods and setting them equal yielding

4! ((kg

2 + l12 ) / l1g) = 4! ((kg

2 + l22 ) / l2g)

A little rearrangement yields

kg

2 + l12 = (kg

2 + l22 )l1 / l2

kg

2 ((l2 ! l1) / l2 ) = l1(l2 ! l1) or

kg

2 = l1l2 . Therefore the period is

T = 2! (l1l2 + l12 ) / l1g = 2! (l2 + l1) / g

Hence

g = 4! 2 (l2 + l1) / T 2 . The answer only depends on the distance between the knife-edges l1 + l2 , and the measured period, T .

W13D3-3 Group Problem Small Oscillations Lennard Jones Potential Solution A commonly used potential energy function to describe the interaction between two atoms is the Lennard-Jones 6,12 potential

U (r) =U0 (r0 / r)12 ! 2(r0 / r)6"# $% ; r > 0 .

where r is the distance between the atoms. Let m denote the effective mass of the system of two atoms.

a) Determine the value of r such that the potential energy is minimum. b) Determine the value of the second derivative of the potential energy at the minimum

of the potential.

c) Find the angular frequency of small oscillations about the stable equilibrium position for two identical atoms bound to each other by the Lennard-Jones interaction.

Solution: The equilibrium points are found by setting the first derivative of the potential energy equal to zero,

0 =

dUdr

=U0 !12r012r!13 +12r0

6r!7"# $% .

Therefore the equilibrium point is located at r = r0 . The second derivative of the potential energy is

d 2Udr 2 =U0 +(12)(13)r0

12r!14 ! (12)(7)r06r!8"# $% .

Evaluating this at r = r0 yields

d 2Udr 2 (r0 ) = 72U0r0

!2

The angular frequency of small oscillation is therefore

!0 =

d 2Udr 2 (r0 ) / m = 72U0 / mr0

2 .