12. For each of the possible wave forms below, indicate which satisfy the wave
equation, and which represent reasonable waveforms for actual waves on a string. For
those which do represent waves, find the speed and direction of propagation, and
sketch the waveform at t=0. The quantities a, b, c, A, and v are all positive constants.
a. yx, t   ax  bt  c 
(4 pts.)
b. y x, t  
(4 pts.)
x t
 
a b
(4 pts.)
ax  b
c. y x, t   A sin 
d. yx, t   A sin ax 2  bt 2
(4 pts.)
a) y(x,t) = (ax + bt + c)2
This function can solve the wave equation, so in that sense it is a solution. However,
it is a parabola which goes to infinity at large x, so it does not describe a reasonable
b) y(x,t) = 1/(ax2+b)
This function has no time dependence, so cannot be a wave.
c) y(x,t) = Asin(x/a + t/b)
This can be rewritten y(x,t) = Asin( 1/a (x + at/b)). It is of the form f(x-vt) with v =
-a/b. This is a reasonable wave form. It travels in the –x direction with speed a/b.
d) y(x,t) = Asin(ax2 – bt2)
This cannot be written in the form f(x-vt). Also, if you plug it in the wave equation
it does not give a constant velocity at all points in space and time.
Grader: 5 pts each. 3 for successfully plugging in, and 2 for explaining if it’s
a good wave or not. (The first 2 points are for cases where partial credit is
needed, some of these can be solve/explained without plugging into the
wave equation).
13. (20 pts) It is observed that a pulse requires 0.1 second to travel from one end to
the other of a long string. The tension in the string is provided by passing the string
over a pulley to a weight which has 100 times the mass of the spring.
(a) What is the length of the string?
(b) What is the equation of the third normal mode?
100  L g
L  vt 
Plug in an expression for the velocity: v  T  where the tension is
equal to the weight of the mass: 100Lg :
The mass density cancels:
L  100Lg t
Solve for L:
L  100g t 
L  9.8m
Grader: 10 pts. They may plug in 0.1 seconds right away and have
strange looking expressions with incorrect units, but that is OK if they
get the right answer.
This is a clamped string for which we have written the normal modes
before. Simply plug in 3 for n:
 3x 
y  An sin
 cos n t 
 L 
Since we have some information on the velocity, the n can be given
 3x 
y  An sin
 cosvkn t 
 L 
The velocity was simplified above:
 3x 
y  An sin
 cos 10 Lg k n t
 L 
Use the standard definition of the wavenumber:
 3x  
y  An sin
 cos10 Lg
 L  
t 
And the expression for the normal mode wavelengths for a clamped string:
2n 
 3x  
y  An sin
 cos10 Lg
2L 
 L  
 3x  
y  An sin
3t 
 cos10
 L  
g 
 3x  
y  An sin
 cos 30
L 
 L  
For this problem, we found above that g and L are numerically equal, so
you can express it like this:
 3x 
y  An sin
 cos30t 
 L 
Grader: 10 pts. If they left it in terms of g and L to avoid an apparent
problems with units, give full credit.
14. Consider a string extending form x = 0 to x = L as shown below. At x = L the
string is fixed to a rigid wall, and at x = 0 the string is attached to a ring of negligible
mass which is free to slide on a friction-free rod. (friction free, so no transverse
a) Using the appropriate boundary conditions, give the solution yn(x,t) for the nth
normal mode, and give explicitly the frequency n and the wavelength n.
(20 pts.)
b) Sketch the patterns for the three lowest allowed modes. (10 pts.)
Hints: Think carefully about the x=0 boundary condition. The size of the rod and massless
ring are negligible. Perhaps a free body diagram would be in order!
----------------------------The right boundary condition is simply that for a clamped string: the y-value is zero
at all times.
yL, t   0
For the left boundary condition, consider a free body diagram of the massless ring:
The massless ring feels a horizontal force from the rod which opposes the
horizontal component of the string tension. The lack of friction means that there
will be no vertical force from the rod. Since the ring is massless, there must also be
no vertical force from the string. If there were, the massless ring would instantly
respond (with infinite acceleration) and move to a point where the vertical
component of the tension is zero. Since the vertical tension is proportional to the
slope of the string (dy/dx), the slope must be zero. So the boundary condition is:
x x  0
Just as initial conditions can apply to the function (initial position) or its first
derivative (initial velocity), boundary conditions can apply to the function (constant
position) or its first derivative (constant slope).
How you reach the answer depends on your initial guess for the string shape. Since
the left side is zero and the origin is displaced, cosine would be a good choice:
yx, t   A coskx   cost 
Apply the left boundary condition:
  Ak sin  cost   0
x x0
which tells us that  = 0.
Now apply the right boundary condition:
yL, t   A coskLcost   0
which is only true if:
kn L 
km L 
2m  1
n  1,3,5,7,...
m  1,2,3,4,5,....
Since k = 2/(always!), the normal mode wavelengths are:
n 
n  1,3,5,7,...
Since  = v k(always!), the normal mode frequencies are:
n 
n  1,3,5,7,...
The normal mode solutions are therefore:
 n   nv 
y n  A cos
x  cos
 2L   2L 
n  1,3,5,7,...
Here are the patterns for the first three modes:
Grader: There are many ways to do this problem! Correct identification and
use of the two boundary conditions is worth 10 points. The boundary
condition at x = L can only be as I stated it. For the boundary condition at x
= 0, they may assume that the end will oscillate with amplitude A and get the
same answer. Note that they may also neglect the phase if they choose
wisely between sine and cosine to describe the shape (some will understand
this, and some will get lucky).
The expression for wavelengths is worth 5 pts, frequencies are worth 5 pts,
and normal modes are worth 4pts. Note that these expressions may vary
depending on choice of sine or cosine, and on choice of odd integers or all
The sketches are worth 2 pts each.
15. (20 pts) A stretched string of mass m, length L, and tension T is driven by two
sources, one at each end. The sources both have the same frequency  and
amplitude A, but are exactly 180 degrees out of phase with respect to one another.
(Each end is an antinode). What is the smallest frequency  consistent with
stationary vibrations of the string (a normal mode)? This is French problem 6-5. (20
The stationary state with smallest frequency will be the first normal mode. Since
both ends are anti-nodes, the n=1 normal mode has a wavelength equal to twice the
string length L:
A cost 
Acost   
1  2L
You can also get this result mathematically by guessing a solution:
yn x, t   An coskn x cosnt 
Its best to guess cosine since the string has amplitude at x = 0. Apply the left
boundary condition:
yn 0, t   An cos0cosnt   A cost 
This tells us that the amplitude of the normal mode (An) must just be the amplitude
of the drive oscillation (A). Also, the frequency of the normal mode must be the
frequency of the drive oscillation.
Now for the right boundary condition:
y n L, t   A cosk n Lcost   A cost   
Here I have plugged in L at the right boundary, and set it equal to the value of the
left boundary, but 180 degrees out of phase.
This is true if:
cosk n L   1
which is true for
k n L  n
odd n (1,3,5,7,9,...)
Or you can handle the odd n this way:
k n L  2n  1
Now to get the frequency:
L  n
n 
all n (1,2,3,4,5,6,…).
odd n (1,3,5,7,9,...)
The lowest frequency will be n=1:
1 
In terms of the parameters given:
1 
 TL
1  
Grader: 10 pts for getting that the first normal mode wavelength is 2L (which
can be done with equations or graphically), and 10 pts for finding the lowest
frequency. If they stop too early and leave the expression for lowest
frequency in terms of v (velocity) rather than T and m, take 2 pts.