Assignment 5, solutions

Problem 1:

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:

Problem 2:

imageTwo 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 m1 is given a displacement L to the right, while m2 is held fixed.  At t = 0 they are released from rest.  Give the equation for the positions of m1 and m2 as a function of time.

Solution:

Problem 3:

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:

Problem 4:

A space scientist proposes to measure the Gravitational constant G by locating a solid gold sphere of mass 8 x 104 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 Ff = -bv), and the change in period caused by the frictional force.

Solution

Problem 5:

image 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:

(b)image  1.)  ω1 = 0, A1 = A2 = A3 = A4 = A5 = A6, = A7 = A8,  ∑jkij = 2k - 2k = 0.
All other modes must have zero total or net angular momentum.

image2.)  Assume all masses M are fixed.  A1 = A3 = A5 = A7 = 0.
The net force on each of these masses must be zero.
For the light masses we then have A2 = A6,  A4 = A8,  A2 = -A4.
To solve for ω22 we can use the equation of motion for mass 2.
k22A2 - ω22m A2 = 0,  2k - ω22m  = 0.
ω22 = 2k/m.

image3.)  Assume all masses m are fixed.  A2 = A4 = A6 = A8 = 0.
The net force on each of these masses must be zero.
For the heavy masses we then have A1 = A5,  A3 = A7,  A1 = - A3.
To solve for ω32 we use the equation of motion for mass 3.
k33A3 - ω32M A3 = 0,
ω32 = 2k/M.

image4.)  Assume only m1 and m5 are fixed.  A1 = A5 = 0.
How can this happen?  The net force acting on m1 and m5 must be zero.
A2 = A4,  A3 = aA2,  A6 = A8 = -A2,  A7 = -A3.
We can use the equations of motion for masses 1 and 3 to solve for ω42 and a.
mass 2:  k22A2 + k23A3 - mω2A2 = 0,  or  2kA2 - kaA2 - mω2A2 = 0,  2k - ka - mω2 = 0.
mass 3:  k32A2 + k33A3 + k34A4 - mω2A2 = 0,  or  -2kA2 + 2kaA2 - Mω2aA2 = 0,  -2k2 + 2kaA - Mω2aA = 0.
Eliminate ω2 and solve for a.  a4 = m/M ± ((M2 + m2)/M2)½.
Choose a4 = 1 - m/M + ((M2 + m2)/M2)½.  a4 > 0.
Insert to solve for ω42
ω42  = k/m + k/M - k(M2 + m2)½/(mM).

image5.)  Assume m1 and m5 are fixed.  A1 = A5 = 0.
A2 = A4,  A3 = aA2,  A6 = A8 = -A2,  A7 = -A3.
Choose a5 = m/M - ((M2 + m2)/M2)½.  a5 <  0.
ω52 = k/m + k/M + k(M2 + m2)½/(mM).

image6.)  Assume m2 and m6 are fixed.  A2 = A6= 0.
A3 = A5,  A4 = aA3,  A7 = A1 = -A3,  A8 = -A4.
(We are just relabeling i --> i + 1 and interchanging m an M in case 4.
a6 = 1 - M/m + ((M2 + m2)/m2)½.  a6 > 0.
ω62 = k/m + k/M - k(M2 + m2)½/(mM).
Note:  mode 4 and 6 are degenerate.  Any linear combination of those modes is also a normal mode.

image7.)  Assume m2 and m6 are fixed.  A2 = A6= 0.
A3 = A5,  A4 = aA3,  A7 = A1 = -A3,  A8 = -A4.
Choose 7 = 1 - M/m - ((M2 + m2)/m2)½.  a7 < 0.
ω72 = k/m + k/M + k(M2 + m2)½/(mM).
Note:  mode 5 and 7 are degenerate.  Any linear combination of those modes is also a normal mode.

image8.)   Assume none of the masses are fixed.
A1 = A3 = A5 = A7.   A2 = A4, = A6 = A8.  A2 = aA1.
All heavy masses oscillate against all light masses.
mass 1:  k11A1 + k12aA1 + k18aA1 - Mω2A1 = 0,  2k -  2ka - Mω2 = 0. 
mass 2:  k21A1 + k22aA1 + k23A1 - mω2aA1 = 0,  -2k +  2ka - maω2 = 0.
Solve for a.  a8 = -M/m.
Solve for ω82.  
ω82 = 2k/m + 2k/M
 

(c)  See diagrams.