Chapter 15a: Mechanical Waves Hints1- Updated 9/19/10
A mechanical wave is a disturbance that travels through a substance called a medium. The wave is
propagated by successive oscillations of the particles in the medium. If the oscillations are small their
motion is simple harmonic and the wave transmits energy through the medium at a constant speed*
without a net transfer of matter. Wave phenomena are very common, and it forms the basis of most of
our communication mechanisms.
*This is not strictly true, but is a good approximation for waves with small amplitudes.
1. Explain the difference between transverse and longitudinal waves, and consider the following
questions. Justify your reasoning. Refer to your notes. Most of the answers below are somewhat
conditional, there may be exceptions…
a) Is it possible to have a longitudinal wave in a taut string?... in a stretched spring?
Longitudinal waves are made up of compressions and rarefactions (stretches), so a medium musts be
compressible to carry longitudinal waves. Strings cannot be compressed so they can’t carry them, but
springs can be compressed and therefore can carry them.
b) Is it possible to have a transverse wave in a steel rod?…in a liquid?
Transverse waves need a rigid medium that can be deformed perpendicular to the wave. A rigid rod
may be able to carry them if it’s thin enough to flex. Liquids have difficulty carrying transverse waves,
but the surface can undulate transversely as bits of surface fluid more around in circles (recall
video…).
c) In an earthquake two kinds of waves are generated S (shear-transverse) and P (pressurelongitudinal). The S waves travel through the earth more slowly than the P waves (approx. 5 km/s vs.
9 km/s). How can detecting the time of arrival of the waves determine how far away was the epicenter
? How many detectors are necessary to pinpoint the epicenter?
The time difference between the signals can determine how far away (d) the epicenter is [∆t=(d/vs)(d/vP)]. But it is difficult to tell the direction of the incoming waves with a single detector, so a second
detector would be required at another location to truly pinpoint the epicenter.
2. Waves come in many types, and there are mathematical equations to describe them. Consider a
single pulse wave. At t=0 this transverse wave pulse in a string is described by the function y=6/(x2
+2).
a) This just looks like a pulse centered at x=0.
b) y(x,t) = 6/[(x - 3t)2 +2].
c) These look like the pulse at the origin but centered at x=3 and x=6 m.
d) y(x,t) = 6/[(x + 3t)2 +2].
3. Two pulses traveling on the same string are described by the following two equations:
y1 = 5/[(3x- 4t)2 + 2]
and
y2 = -5/[(3x+ 4t- 6)2 + 2]
a) y1 travels right and y2 travels left. To find speed, factor “3” out of the squared expression in the
denominator to display (x-vt). The speed is 4/3 m/s assuming m and sec units.
b) 2 m apart; c) x=1 m ; d) ¾ sec.
4. The basic formula for this wave is: y(x,t) = A sin (kx- t + ), where k is the “wave no.” (k=2π/,
is the “angular frequency” (=2πf=2π/T), and  is the phase angle that describes the initial shape of
the wave [y(0,0)=A sin].
a) This is just algebra.
b) The proof is a logical argument: At t=0, y(x,0)= f(x-0). At x=vt, y[(x+vt), t]= f(x-0) again! So
the form of the wave at t=0 is the same at “x=vt” at time “t”. Hence the wave has “traveled”
a distance “x=vt” during time “t”.
c) Same argument but with –v.
5.
The “trick” with these is to redraw the wave a short time afterwards as it moves in the direction of v.
Since the oscillators are fixed in the x-axis, they have to move up or down (or back & forth) to reshape
the wave as it travels.
In a, b, and c, the leading part of the crests and trailing part of the troughs are moving up, while the
other parts are moving down. The equilibrium points are moving the fastest, while the max and min
are instantaneously at rest.
a)
y
b)
v
y
x
c)
v
y
d)
v
x
x
In (d), the oscillations are “back & forth” parallel to the speed of the wave, so the compressions are
instantaneously moving right and the rarefactions are instantaneously moving left. The in-between
points are instantaneously at rest. See your notes to review the logic.
e) Sketch rough graphs of the instantaneous speeds of the oscillators vs. “x” [(dy/dt) vs. x] for the
cases above. Note that these are NOT the slopes of the graphs above because these are “y vs. x”
graphs, not “y vs. t” graphs. …Coming later…
6. Simply take derivates of y(x,t) = A sin (kx- t + ):
vy=dy/dt= (-Acos (kx- t + ); max vy= ±Aat the equilibrium position; min vy= 0, at the extreme
positions.
ay=dvy /dt)= (-A) sin (kx- t + ); max ay= ±Aat the extreme positions; min ay= 0, at the
equilibrium position.
The following 5 problems are designed to give you practice working with the sinusoidal traveling wave
function. There are many variations of these problems since the given information can vary. You don’t
have to do all of these but you will be expected to do the following things:
1. Derive important information from a given equation.
2. Write the wave function given basic information about the wave.
3. Sketch snapshots of the wave (y vs. x) at different times, and time graphs (y vs. t) for points at
different x locations.
4. Determine instantaneous velocities and accelerations (vy and ay) for the individual oscillators
in the medium.
7A. A sinusoidal wave of =2 m and A=0.1 m travels with v=1 m/s on a string. Initially the left end of
the string is at the origin and waves moves from left to right. Determine:
a) f= 0.5 Hz; rad/sec; k= π m; b) y(x,t)=0.1 sin (πx- t):
c) y(x, 0)= 0.1 sin(πx); y(x, T/4)= 0.1 sin(πx- π/2)= 0.1 cos(πx); y(x, T/2)= 0.1 sin(πx- π)= -0.1 sin(πx).
You know what these look like…
d) y(0, t)= 0.1 sin(-t)= -0.1 sin(t);
e) y(1.5, t)= 0.1 sin(1.5π-t)= -0.1 cos(t)
y
7B. The illustration shows a traveling wave at y(x,0) and at y(x,0.2s)
with a speed of 1.25 m/s to the right. Determine
a) A=6 cm; f=1.25 Hz; m.
b) y(x,t)=0.06 cos (2πx- 2.5t)= 0.06 sin (2πx- 2.5t +π/2)
c) y(0,t)= 0.06 cos(- 2.5t)= 0.06 cos(2.5t) ;
x=0, t=0, y=6cm
x
t=0.2s
later
y(1.5, t)= 0.06 cos(3π - 2.5t)= -0.06 cos(2.5t); The phase difference is π.
d) vy=dy/dt= (-0.15πsin(0.5π -0.5 ) =0
7C. Consider this wave function: y(x,t) = 0.35 sin(10πt - 3πx + π/4), using cm and sec units. What
is…
a) v=+10/3=3.3 cm/s to the right
b) y(x,t) = 0.35 sin(-30π + π/4)= 0.35 (0.707) =0.25cm
c) Max vy= Amax ay= A =3.45 m/s2
d) Since the max acceleration of the max acceleration of the string is less than that of gravity (9.8
m/s2), the ant will not be knocked off the string by the oscillations, it will just feel lighter when the
string accelerates down.
7D. For the wave illustrated in the diagram: a) Find the period and amplitude. b) The points x=0 and
75 mm are within one wavelength of each other and the wave is moving in +x direction. Find  and v.
c) If instead the wave is moving in the –x direction, what is  and v? Justify your answers.
d) If you don’t know that the points are within a wavelength apart, would your answer for the
x=0,
x=75mm,
wavelength be unique?
y(mm)
t= 10ms,
y=4mm
t=35ms,
y=4mm
x=0,
a) Study the info given for the peaks. From the graph A=4 mm
t=50ms,
and ¼T=10 ms, so T=40 ms.
y=4mm
t(s)
b) The peak at x=0 at 10 m) travels to x=75 mm at 35 ms.
So v=75mm/25ms= /T  v=3 m/s and=120 mm.
Or you can use the wave formula: y(0,10)=y(75,35) -20π/40=(150π/ -70π/40) and solve for .
c) Here v=75mm/15ms= /T  v=5 m/sand  =200 mm. The reason is that the signals now move
from the 75 mm point to the 0 point since the wave travels in the –x direction. The peak at x=75 mm at
35 ms travels to x=0 at 50 ms. Or you can substitute into the equations since y(75,35)= y(0,50)…
d) No, because the time graphs would look the same after every cycle…
7E. A transverse sinusoidal wave on a string has a period T= 0.25 s and travels in the –x direction with
a speed of 30 m/s. At t=0, a particle on the string at x=0 is displaced vertically +2 cm and has an
instantaneous vertical speed of +2 m/s.
a) Combine y(0,0)= A sin = 0.02 m, and vy(0,0) =A cos =2 m/sA= 0.082 m
b) From the same formulas in (a)   =0.25 rad = 14º
c) A = 0.082(8π)=2.1 m/s
x=0, t=0,
d) y(x,t)=0.082 sin [(2πx/7.5)+ 8t +0.25]
y=2cm
e) We want y(0,t)=0.082 So 1=sin(8t +0.25)π/2=8t +0.25 t=0.053 s ~1/5 of a period
9. It can be proven that the speed of a transverse wave in a string is given by the formula: v=√(FT/µ),
where FT is the tension in the string and µ is the linear mass density (m/L) of the string.
a) Remember that N=kg(m/s2).
b) 4
c) No change, frequency doesn’t affect the speed.
d) No change, since the density if the string doesn’t change.
e) Realistically, a large amplitude could stretch the string and decrease its density somewhat, so it
could affect the speed.
f) Using the given info: =(2/4.5) m, f=5 Hz (note that f is not 6 because the first crest is “0” in the
count of cylces), and µ=(0.005/2) kg/m. Plugging in I got F=0.012 N.
8. The average energy transferred by a transverse wave each second is given by the formula: Pave
=µvA2/2. Consider a case where transverse waves are being generated in a wire under constant
tension. Assume the density of the wire doesn’t change. By what factor must the average power be
increased or decreased if
a) 1 (No change); b) 1 (No change); c) 4
d) No, since the density and the tension remain constant.
e) Plug into formula and watch out for units. Everything needs to be in mks units. I got 0.22 watts.
f) Doubling the cross-sectional area would double µ, but it would also reduce v=√(F/µ) by 1/√2. So
the required power would increase by a factor of √2.
g) Generally the oscillators have some potential and some kinetic energy. But if energy is conserved, the
total is equal to either the max potential energy (when K=0) or the max kinetic energy (when U=0).
10. A whip of density  tapers to nearly a point so that its cross-sectional area A varies along its length
x, according to the formula: A= (10 – 0.1x), where x is in cm and A in cm2.
a) First you need to find an expression for the linear mass density in terms of the volume density and area:
µ=dm/dx = dVol)/dx) = A(dx/dx)=  A=  (10 – 0.1x) v=√[F/ (10 – 0.1x)]
b) Here v2A =F. Plugging in, I got F =1.47 N = 147 kilodynes. It is easier to use “cgs” units (cm,g,
dynes…) in the rest of this problem.
c) According to the given formula, A= 10 cm2 at x=0. Assuming the tension is the same as that found in (b),
and plugging into the velocity formula, I got v=111 cm/s.
d) Set-up the integral dt=dx/v = (/F)1/2 ∫A1/2 dx =(/F)1/2 ∫(10 – 0.1x)1/2 dx , and integrate over the length of
the whip (x=0 to 100 cm). I got t~0.6 sec.
e) Since the speed is getting faster the pulse must stretch out and flatten out as it travels.
11. A rope of total mass m and length L is suspended vertically from the ceiling.
a) In this situation the tension of the rope increases as you go up due to the increasing rope weight
that is supported from the higher points on the rope .
b) Following the hint you can show that v=√(gy), where y is a vertical point on the hanging rope
(measured from the bottom) and g is the acceleration due to gravity. Since v=dy/dt, you can set up the
integral dt=dy/v, then integrate both sides to get the given answer. tL= 2(L/g)1/2
c) tL/2= (2L/g)1/2
d) L/4. These last two questions should convince you that the relationship between y vs. t is not linear.
e) The pulse “stretches out and flattens” as it travels up.
Challenge problems:
12. For a more challenging version of the problem above, assume that there is a mass M hanging from
the bottom of the rope.
a) Show that the time for a pulse to reach the top is t= 2(L/mg)1/2[√(M+m) -√M].
b) You can further show that this reduces to the answer in 11(a) when M=0 and that it reduces to
(mL/Mg)1/2 when m<<M.
a) The method is the same as for problem above, except the hanging weight adds to the weight of the
dangling string to make up the tension: F=(myg/L) +Mg.
b) The case where m<<than M requires the use of the binomial expansion approximation: (1 +x)n~=
1+nx, when x<<1.
Write the term [√(M+m) -√M] as [√M (1-m/M)1/2] and apply the binomial approx…
13. In class we used a long spring to demonstrate longitudinal (as well as transverse) waves.
a) Show that if a spring obeys Hooke’s law (F=kx) and has mass m, length L, and force constant k, the
speed of the longitudinal waves in the spring is given by v=L√(k/m).
b) Evaluate v for a spring of m= 0.250 kg, L= 2 m, and force constant k=1.5 N/m.
a) The derivation is conceptually challenging and requires a careful comparison between the motion
of the oscillators and the advancing “wavefront” and applying the “impulse-change in momentum”
principle.
One starts by stating that during a time ∆t a portion of the spring is compressed an amount ∆s while
the wavefront advances a distance ∆x along the spring. Applying the “impulse-change in momentum”
principle: F∆t=(∆m)vs, where ∆m is the mass of the spring segment ∆x and vs =(∆s/∆t) is the
instantaneous speed of the oscillators due to the impulse. The force F=k’∆s, where k’=kL/∆x (this is
the spring constant of the segment ∆x, since the rest of the spring is not engaged at this stage) and
∆m=µ∆x=m∆x/L. Substituting gives: (kL/∆x)∆s∆t=(m∆x/L)(∆s/∆t). After simplifying and rearranging
one gets kL2=m(∆x/∆t)2. Since (∆x/∆t) is the speed of the wave “v”, we have the expression we were
seeking: v=L√(k/m).
Some textbooks do this proof, so you can look it up. Or you can see me and I’ll discuss the proof with
you.
b) Just plug in….