A train has a whistle, which emits a 400 Hz sound. You are stationary and you hear the whistle, but the pitch is 440 Hz. How fast is train moving towards or away from you?

Solution:

- Concepts:

The Doppler shift - Reasoning:

The frequency of the train's whistle is shifted, because the train is moving with respect to the observer. - Details of the calculation:

The pitch is higher, so the train is moving towards you. The train is a moving source.

Its speed relative to you is found from f = f_{0}v/(v-v_{s}). We have

(v-v_{s}) = f_{0}v/f = (400/s)(340 m/s)/(440/s) = 309 m/s.

Therefore v_{s }= 340 m/s - 309 m/s = 31 m/s = 69 mph.

Two masses, 1 kg and 2 kg, are fixed horizontally to fixed side supports
with springs as shown below. The masses are constrained to move along the horizontal line. From their equilibrium position m_{1} is given a displacement L to
the right, while m_{2} is held fixed. At t = 0 they are
released from rest. Give the equation for the positions of m_{1} and m_{2}
as a function of time.

Solution:

- Concepts:

Coupled oscillations, normal modes - Reasoning:

Small oscillations:

L = ½∑_{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}] with T_{ij}= T_{ji}, k_{ij}= k_{ji}.

Solutions of the form q_{j}= Re(A_{j}e^{iωt}) can be found.

For a particular frequency ω_{α}we solve

∑_{j}[k_{ij}- ω_{α}T_{ij}]A_{jα}= 0

to find the A_{jα}. The most general solution for each coordinate q_{j}is α sum of simple harmonic oscillations in all of the frequencies ω_{α}.

q_{j}= Re∑_{α}(C_{α}A_{jα}exp(iω_{α}t)). - Details of the calculation:

The kinetic energy is T = ½[m_{1}(dx_{1}/dt)^{2}+ m_{2}(dx_{2}/dt)^{2}],

and the potential energy is U = ½)[k_{2}(x_{2}- x_{1})^{2}+ k_{1}x_{1}^{2}+ k_{3}x_{2}^{2}].

k_{1}= 2 N/m, k_{2}= 2 N/m, k_{3}= 6 N/m.

U = ½[(k_{1}+ k_{2})x_{1}^{2}+ (k_{3}+ k_{2})x_{2}^{2}- k_{2}( x_{1}x_{2}+ x_{2}x_{1})].

The x_{i}are the displacements from the equilibrium positions. We use the x_{i}as our generalized coordinates q_{i}.

The Lagrangian is L = T - U. This can be put into the form

L = ½∑_{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}] with T_{ij}= T_{ji}, k_{ij}= k_{ji}.

Here T_{11}= m_{1}= 1, T_{22}= m_{2}= 2, T_{ij}(i≠j) = 0,

k_{11}= 5, k_{22}= 8, k_{12}= k_{21}= -2.

All numerical quantities are in SI units.

|k_{ij}- ω^{2}T_{ij}| = 0, (k_{11}- ω^{2}T_{11})(k_{22}- ω^{2}T_{22}) - k_{12}^{2}= 0, (5 - ω^{2})(8 - 2ω^{2}) - 4 = 0.

ω^{4}- 9ω^{2}+ 18 = 0, ω^{2}= 9/2 ± 3/2. ω_{a}= 6, ω_{b}= 3.

ω_{a}: -A_{1a}- 2A_{2a}= 0, A_{1a}= -2A_{2a}. Choose A_{1a}= 2, A_{2a}= -1.

ω_{a}: 2A_{1b}- 2A_{2b}= 0, A_{1b}= A_{2b}. Choose A_{1b}= A_{2b}= 1.

Most general solution:

x_{1}= 2C_{a}cos(ω_{a}t + φ_{a}) + C_{b}cos(ω_{b}t + φ_{b}), x_{2}= -C_{a}cos(ω_{a}t + φ_{a}) + C_{b}cos(ω_{b}t + φ_{b}).

dx_{1}/dt = -2ω_{a}C_{a}sin(ω_{a}t + φ_{a}) - ω_{b}C_{b}sins(ω_{b}t + φ_{b}),

dx_{2}/dt = ω_{a}C_{a}sin(ω_{a}t + φ_{a}) - ω_{b}C_{b}sin(ω_{b}t + φ_{b}).

Initial conditions at t = 0:

L = 2C_{a}cos(φ_{a}) + C_{b}cos(φ_{b}), 0 = -C_{a}cos(φ_{a}) + C_{b}cos(φ_{b}). L = 3C_{a}cos(φ_{a})

0 = -2ω_{a}C_{a}sin(φ_{a}) - ω_{b}C_{b}sins(φ_{b}), 0 = ω_{a}C_{a}sin(φ_{a}) - ω_{b}C_{b}sin(φ_{b}). φ_{a}= φ_{b}= 0.

C_{a}= C_{b}= L/3.

Positions of m_{1}and m_{2}as a function of time:

x_{1}= (2L/3)cos(√6 t) + (L/3)cos(√3 t), x_{2}= -(L/3)cos(√6 t) + (L/3)cos(√3 t).

Consider a "molecule" made up of three equal masses m connected by three equal springs with spring constant k. The equilibrium position is an equilateral triangle. Consider only motion in the plane of this triangle. Find all the normal modes for motion in this plane.

Solution:

- Concepts:

Coupled oscillations, normal modes, symmetry - Reasoning:

We are asked to find the normal modes of a system of 3 particles connected by springs. The particles move in a plane.

m_{1}= m_{2}= m_{3}= m.

We have three masses moving in two dimensions, so there are six normal modes.

Three of of these modes will have ω = 0. Two of these correspond to uniform translation of the CM along two perpendicular axes and all masses stay at their equilibrium position. The other ω = 0 mode corresponds to uniform rotation about the CM. The springs are stretched by an amount 2qcos(30^{o}), so that 4kqcos^{2}(30^{o}) = 3kq = mrΩ^{2}. Ω is the angular speed of the uniform rotation.

If the equilibrium length of the springs is L then r = L/(2cos30^{o}) + q = (2/3)Lcos30^{o}+ q.

The remaining modes keep the CM fixed and have no angular momentum about the CM.

One of the remaining modes is the "breathing mode".

Each mass moves along the line joining the CM and its equilibrium position. The masses move in and out in unison.

T = (3/2)m(dq/dt)^{2}, U = (3/2)k(2qcos(30^{o}))^{2}= (9/2)kq^{2}, L = T - U.

Equation of motion. 3md^{2}q/dt^{2}+ 9kq = 0, ω^{2}= 3k/m. - Details of the calculation:

The remaining two modes are degenerate.

Let us look at one of these. When m_{1}and m_{2}both move up a distance χ, m_{3}moves down a distance 2χ. At the same time m_{1}moves a distance -η and m_{2}moves a distance η along the x-direction.

T = 2*½m[(dη/dt)^{2}+ (dχ/dt)^{2})] + ½m(d2χ/dt)^{2}= 3m(dχ/dt)^{2}+ m(dη/dt)^{2}.

U = ½k(2η)^{2}+ 2*½kΔ^{2}. Δ = L' - L.

L'^{2}= (Lcos30^{o}- 3χ)^{2}+ (L/2 + η)^{2}= (L + Δ)^{2}.

Assuming small displacements and keeping only terms to first order in the small quantities we have

-3χcos30^{o}+ η/2 = Δ. -(3√3/2)χ + η/2 = Δ.

U = 2kη^{2}+ k(27χ^{2}/4 + η^{2}/4 - (3√3/2)χη) = ½[(9/2)kη^{2}+ (27/2)kχ^{2}- (3√3/2)k(χη + ηχ)].

T = ½∑_{ij}T_{ij}(dq_{i}/dt)(dq_{j}/dt), U = ½∑_{ij}k_{ij}q_{i}q_{j}, with q_{1}= η and q_{2}= χ.

T_{11}= 2m, T_{22}= 6m, T_{ij}(i≠j) = 0, k_{11}= 9k/2, k_{22}= 27k/2, k_{12}= k_{21}= -(3√3/2)k.

det(k_{ij}-ω^{2}T_{ij}) = 0.

(9k/2 - 2mω^{2})(27k/2 - 6mω^{2}) - 27k^{2}/4 = 0. ω^{4}- 4.5(k/m)ω^{2}+ 4.5(k/m)^{2}= 0.

ω^{2}= 3 k/m or ω^{2}= 1.5 k/m.

We rediscover the "breathing mode", with ω^{2}= 3 k/m, χ = -η/√3.

For ω^{2}= 1.5 k/m we have (9/2 - 3)η - (3√3/2)χ = 0, χ = η/√3.

As masses m_{1}and m_{2}move up a distance χ, there horizontal separation increases by 2η = 2√3x.

Another of the degenerate modes has masses m_{2}an m_{3}moving towards the CM distance χ, while m_{1}moves a distance 2χ towards the CM from the opposite direction. The mode where m_{3}and m_{1}move in unison is a linear combination of the previous two independent degenerate modes.

A space scientist proposes to measure the Gravitational constant G by
locating a solid gold sphere of mass 8 x 10^{4} kg and radius 1 m in a
spaceship. A hole is to be drilled through the diameter of the sphere and
a small gold ball of mass 80 gram is released from rest at its surface so that it
oscillates back and forth within the tube passing through the large sphere.

(a) Calculate the period of the oscillatory motion of the particle,
assuming that there is no friction present.

(b) If air friction is
present so that the particle returns to a radial distance of 99 cm rather than
to 100 cm at the end of one full oscillation, obtain an estimate of the
frictional force (assumed to be of form **F**_{f} = -b**v**), and
the change in period caused by the frictional force.

Solution

- Concepts:

Newton's law of gravity, Newton's 2^{nd}law, the principle of superposition, the underdamped harmonic oscillator - Reasoning:

In part (a) the acceleration of the of the objects is due to the gravitational force. The gravitational force on a object in the hole, a distance r from the center is in the -**r**direction and its magnitude is found using "Gauss' law". [(∇∙F_{g}/m) = -4πGr, for a spherical mass distribution. 4πr^{2}(F_{g}/m) = 4πGM_{inside}, (F_{g}/m) = GM_{inside}/r^{2}.]

In part (b) the oscillator is underdamped, since it crosses the equilibrium position many times. - Details of the calculation:

(a) For an object of mass m in the hole F = (4/3)Gmπρr , with ρ = 3M/(4πR^{3}). M and R are the mass and radius of the large sphere, respectively.

For an object moving in the hole we therefore have**F**= -k**r**, k = (4/3)Gmπρ.

The force on the object obeys Hooke's law. The object will oscillate with angular frequency

ω = (k/m)^{½}= ((4/3)Gπρ)^{½}= 2.31*10^{-3}/s.

Its period is T = 2π/ω = (3π/Gρ)^{½}= 2π(R^{3}/GM)^{½}= 2720 s.

(b) The equation of motion for the damped harmonic oscillator is

d^{2}x/dt^{2}+ 2bdx/dt + ω_{0}^{2}x = 0.

The solution for underdamped motion is x(t) = A exp(-bt) cos(ω_{1}t - δ), with ω_{1}^{2}= ω_{0}^{2}- b^{2}. Here ω_{0}is the angular frequency of an undamped oscillator with the same spring constant. [We define T_{1}as the time between adjacent zero crossings, 2T_{1}as the "period", and ω_{1}= 2π/(2T_{1}) as the "angular frequency" of the underdamped oscillator.]

After one period 2T_{1}the amplitude decreases from 1 m to 0.99 m.

exp(-b2T_{1}) = 0.99, -b2T_{1}= ln(0.99), b2T_{1}= 0.01,

ω_{1}^{2}= ω_{0}^{2}- b^{2}, (2π)^{2}/(2T_{1})^{2}= ω_{0}^{2}- b^{2}, (2π)^{2}b^{2}/(ω_{0}^{2}- b^{2}) = 10^{-4}.

b = 1.59*10^{-3}ω_{0}= 3.68*10^{-6}/s.

(2T_{1 }- T_{0})/T_{0}= (0.01/(1.59*10^{-3}*2π) - 1) = 9.7*10^{-4}.

ω_{1}^{2}≈ ω_{0}^{2}, 2T_{1}≈ T_{0}.

The period of the underdamped motion is practically equal to the period of the undamped motion since b << ω_{0}.

A set of coupled masses is constrained to move on a circular path of fixed
radius. The chain consists of four light masses m alternating with four
heavy masses M, joined by identical springs with force constant k.

(a) Find the (coupled) equations of motion for the ith light mass m and the ith
heavy mass M.

(b) Solve these equations of motion for the normal modes of the system.

(c) Sketch the motion associated with each of the allowed frequencies

Solution:

- Concepts:

Coupled oscillations - Reasoning:

Let q_{i}denote the displacement from equilibrium. Let a clockwise displacement be a positive displacement.

(a) L = ½∑_{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}] with T_{ij}= T_{ji}, k_{ij}= k_{ji}.

Here T_{ii}= m_{i}, T_{ij}(i ≠ j) = 0, i, j = 1, 2, 3, 4, 5, 6, 7, 8, (cyclic).

m_{i}= M if i = odd, m_{i}= m if i = even.

k_{ii}= 2k, k_{ij}= -k if j = i ± 1, k_{ij}= 0 otherwise.

The equations of motions are ∑_{j}(T_{ij }∂^{2}q_{j}/∂t^{2 }+ k_{ij}q_{j}) = 0.

Solutions of the form q_{j}= Re(A_{j}e^{iω t}) can be found. For a system with n degrees of freedom, n characteristic frequencies ω_{α}can be found. Some frequencies may be degenerate. For a particular frequency ω_{α}we solve

∑_{j}[k_{ij}- ω_{α}^{2}T_{ij}]A_{jα}= 0

to find the A_{jα}. Since it is difficult to evaluate the determinant of a 8 x 8 matrix, we find the solutions of this system of coupled equations using physical insight. We need to find 8 normal modes. - Details of the calculation:

(b) 1.) ω_{1} = 0, A_{1} = A_{2} = A_{3} = A_{4}
= A_{5} = A_{6}, = A_{7} = A_{8}, ∑_{j}k_{ij} = 2k -
2k = 0.

All other modes must have zero total or net angular momentum.

2.) Assume all masses M are fixed. A_{1} = A_{3}
= A_{5} = A_{7} = 0.

The net force on each of these masses must be zero.

For the light masses we then have A_{2} = A_{6},
A_{4} = A_{8}, A_{2} = -A_{4}.

To solve for ω_{2}^{2} we can use the equation
of motion for mass 2.

k_{22}A_{2} - ω_{2}^{2}m_{
}A_{2}
= 0, 2k - ω_{2}^{2}m_{ } = 0.

ω_{2}^{2} = 2k/m.

3.) Assume all masses m are fixed. A_{2} = A_{4}
= A_{6} = A_{8} = 0.

The net force on each of these masses must be zero.

For the heavy masses we then have A_{1} = A_{5},
A_{3} = A_{7}, A_{1} = - A_{3}.

To solve for ω_{3}^{2} we use the equation of
motion for mass 3.

k_{33}A_{3} - ω_{3}^{2}M_{
}A_{3}
= 0,

ω_{3}^{2} = 2k/M.

4.) Assume
only m_{1} and m_{5} are fixed. A_{1} = A_{5}
= 0.

How can this happen? The net force acting on m_{1}
and m_{5} must be zero.

A_{2} = A_{4}, A_{3} = aA_{2},
A_{6} = A_{8} = -A_{2}, A_{7}
= -A_{3}.

We can use the equations of motion for masses 1 and 3 to solve
for ω_{4}^{2} and a.

mass 2: k_{22}A_{2} + k_{23}A_{3}
- mω^{2}A_{2} = 0, or 2kA_{2}
- kaA_{2} - mω^{2}A_{2} = 0, 2k -
ka - mω^{2} = 0.

mass 3: k_{32}A_{2} + k_{33}A_{3}
+ k_{34}A_{4} - mω^{2}A_{2} = 0,
or -2kA_{2} + 2kaA_{2} - Mω^{2}aA_{2}
= 0, -2k_{2} + 2kaA - Mω^{2}aA = 0.

Eliminate ω^{2} and solve for a. a_{4} = m/M ± ((M^{2}
+ m^{2})/M^{2})^{½}.

Choose a_{4} = 1 - m/M + ((M^{2} + m^{2})/M^{2})^{½}.
a_{4} > 0.

Insert to solve for ω_{4}^{2}.

ω_{4}^{2}
= k/m + k/M - k(M^{2} + m^{2})^{½}/(mM).

5.) Assume m_{1} and m_{5} are fixed. A_{1} = A_{5}
= 0.

A_{2} = A_{4}, A_{3} = aA_{2},
A_{6} = A_{8} = -A_{2}, A_{7}
= -A_{3}.

Choose a_{5} = m/M - ((M^{2} + m^{2})/M^{2})^{½}.
a_{5} < 0.

ω_{5}^{2} = k/m + k/M + k(M^{2} + m^{2})^{½}/(mM).

6.) Assume m_{2} and m_{6} are fixed. A_{2} = A_{6}=
0.

A_{3} = A_{5}, A_{4} = aA_{3},
A_{7} = A_{1} = -A_{3}, A_{8}
= -A_{4}.

(We are just relabeling i --> i + 1 and interchanging m an M in
case 4.

a_{6} = 1 - M/m + ((M^{2} + m^{2})/m^{2})^{½}.
a_{6} > 0.

ω_{6}^{2} = k/m + k/M - k(M^{2}
+ m^{2})^{½}/(mM).

Note: mode 4 and 6 are degenerate. Any linear
combination of those modes is also a normal mode.

7.) Assume m_{2} and m_{6} are fixed. A_{2} = A_{6}=
0.

A_{3} = A_{5}, A_{4} = aA_{3},
A_{7} = A_{1} = -A_{3}, A_{8}
= -A_{4}.

Choose 7 = 1 - M/m - ((M^{2} + m^{2})/m^{2})^{½}.
a_{7} < 0.

ω_{7}^{2} = k/m + k/M + k(M^{2}
+ m^{2})^{½}/(mM).

Note: mode 5 and 7 are degenerate. Any linear
combination of those modes is also a normal mode.

8.)
Assume none of the masses are fixed.

A_{1} = A_{3} = A_{5} = A_{7}. A_{2} = A_{4},
= A_{6} = A_{8}. A_{2} = aA_{1}.

All heavy masses oscillate against all light masses.

mass 1: k_{11}A_{1} + k_{12}aA_{1}
+ k_{18}aA_{1} - Mω^{2}A_{1} =
0, 2k - 2ka - Mω^{2} = 0.

mass 2: k_{21}A_{1} + k_{22}aA_{1}
+ k_{23}A_{1} - mω^{2}aA_{1} =
0, -2k + 2ka - maω^{2} = 0.

Solve for a. a_{8} = -M/m.

Solve for ω_{8}^{2}.

ω_{8}^{2}
= 2k/m + 2k/M

(c) See diagrams.