# Sound and wave motion I

## © 2013 by  Fernando Caracena

The physics of sound is of broad interest, especially for musicians, and is a good introduction to the science of wave motion, which is an important addition to the kinematics of material objects.

# Standing wave pattern

Imagine children playing jump rope as you are looking at the rope broadside. Someone else is looking straight down from a balcony. What you see is the rope going up and down between two relatively fixed ends in simple harmonic motion (SHM). The other person sees the rope going back and forth, also in SHM. A previous blog has touched on SHM. The rotation of the rope is a combination of two SHMs. Instead of rotating the rope, the children could have caused it to move up and down only, in which case the person on the balcony would see a straight rope with no back and forth motion. This oscillatory motion in a plane between two fixed points is called a standing wave. Except for nodal points where the rope does not move, the other parts of the rope execute SHM with variable amplitudes in the form of a sine wave.

## Simple harmonic motion

A mass (m) held in an equilibrium position (along the y axis at y=0) by a perfectly elastic spring executes simple harmonic motion when it is displaced from equilibrium (y) and released, or if given an initial velocity, v0 when it is at its equilibrium position. Any displacement of the mass from equilibrium (y) produces a restoring force(F), which is proportional and opposite to y,

F = - k y,                                                        (1a)

where k is the spring constant.

Using Newtons laws of motion, we can solve for the acceleration of the displaced mass,

m d2y/dt2 = - k y.                                                (1b)

Two real functions, the sine and cosine are know to satisfy (1b). They have the following derivatives:

dsin(ω t)/dt = ω cos(ω t);                                         (2a)

dcos(ω t)/dt =-ω sin(ω t);                                        (2b)

d2sin(ω t)/dt2 = -ω2sin(ω t);                                     (2c)

d2cos(ω t)/dt 2= -ω2cos(ω t);                                   (2d)

where

ω= 2π /T                                                      (2e)

is the angular frequency (measured in radians per second)

and T is the period of oscillation in seconds (not to be confused with the units of time ([T] or seconds). The inverse of the period is called the frequency (Hz, or cycles per second),

ν= 1/T .                                                         (2d)

Note that the frequency and angular frequency are related to each other to within a factor of 2π,

ω= 2π ν.                                                         (2f)

The solution to (1b) is written below in terms of the sine function

y= A sin(ω t + Φ),                                          (3a)

where  Φ specifies the starting phase of the motion. We can show this by using the trigonometric identity,

sin(B+C)= sin(B) cos(C) + cos(B) sin(C),

which reduces (3a) to a sum of sine and cosine components,

y= A sin(ω t) cos(Φ) + A cos(ω t) sin(Φ) .     (4a)

The phase factor allows one to fit the initial conditions. For example,

suppose the mass of the harmonic oscillator is initially at x=0 but is given an initial velocity to the right

dy/dt=ω A [cos(ω t) cos(Φ) - sin(  ω t) sin(Φ)].   (4b)

At t=0, (4a) and (4b) become

y=A sin(Φ),

dy/dt=ω A cos(Φ),

respectively. The initial conditions are satisfied if we set Φ =0, in which case

A= v0/ ω

and

y= A sin(ω t ),

where v0 is the initial velocity of the mass.

Exercise, show that if the mass is released from rest at a displacement, x0

the phase factor is   Φ =π/2 (90 degrees), so that

y= A cos(ω t).

It is possible to solve for the kinematic characteristics of the simple harmonic oscillator described by (1a) and (b) by substituting the solution (3a) for y, in which case we get

-m ω2 y= - k y.

Solving for the angular frequency we have

ω = √(k/m) ,                                          (3b)

or for the frequency,

ν= (2π)-1 √(k/m).                                  (3c)

## Mathematically representing the standing wave pattern

Fig. 1. The fundamental standing wave pattern between tow fixed points along the x-axis.

Here we grok the mathematical description of a standing wave pattern of a rope or string suspended between rigid supports at both ends (x/L=0, 1 or x=0, L), where L is the distance between rigid supports . The string moves up and down parallel to the y-axis, and we are looking for a mathematical function that specifies y as a function of x. The entire description must also specify y as a function of time. We look for a solution of the form

y(x,t)=yn(x) f(t).              (4a)

First, note that the zeroes of the  sine function are given by,

sin(n π x/L)=0,      n= 1, 2, 3, ....∞;                                                                        (4b)

and of the cosine function by,

cos(n π/2 x/L)=0,  n=1, 2, 3, ....∞.                                                                         (4c)

In the case of the standing wave pattern depicted in Fig. 1 we have starting nodes between x=0 and x=L, and several points in between, which are given by the zeroes of  yn as represented as a sine function,

yn(x) = A sin(n π x/L).                                       (5a)

This function (5a) gives the local amplitude of oscillation for each point on the string along the x-axis, which itself is in SHM. All of such points have the same angular frequency, ω. If we chose the phase so that these oscillators all begin at a maximum positive displacement, the phase factor is   Φ =π/2 . In this case, the complete description of the motion of the string is

yn(x,t) = A sin(n π x/L)cos(ω t).                                       (5b)

Although (5b) gives a mathematical description of the shape of the string at any time in the interval x=0, 1, the angular frequency is yet undetermined, which can be solved for in terms of the physical state of the string.

## Restoring force from tension

Fig. 2. Restoring force on an infinitesimal, curved string segment produced by tension.

When a string under tension is subject to a small lateral displacement (in the y-direction), a net force is generated on every curved, infinitesimal section of  the string as a result of the change in orientation of the tension (acting tangentially) across the infinitesimal section (Fig. 2). We analyse this situation by looking at changes in slope across infinitesimal sections of the string in terms of the derivatives of the function that describes the shape of the string, y0(x), where x is the horizontal distance along the string.

The curve shown in Fig. 2 is exaggerated vertically. The tangent to displacement curve is very nearly horizontal and the slope of the tangent is very small compared to unity.

Fig. 3. Trigonometric functional relationships for a small angle.

A right triangle having a small vertex angle (a) is depicted in Fig. 3. The hypotenuse of the triangle is 1, its base is cos(a) and height, sin(a). Note that

sin(a) << 1.

The slope of the hypotenuse is given by a trigonometric identity

tan(a) = sin(a)/cos(a),                                   (6a)

from which we can solve for sin(a)

sin(a) = tan(a) cos(a).                                   (6b)

Tan(a) corresponds to the derivative of the function that specifies the shape of the string, y0(x); but we want to find the value of sin(a), which can be used to find the vertical component of the force exerted by the tension on the infinitesimal string segment.  Below, small angle approximations are used to find sin(a),

sin(a) = tan(a) √(1-sin(a)2),                          (6c)

where we have used the trigonometric identity,

sin(a)2+cos(a)2 =1,

to eliminate the cos(a) from (6a).

when a « 1

sin(a)≈a

Using the following approximation of the square root for small departures from unitiy,

√(1-sin(a)2) ≈ 1-½ sin(a)2

or neglecting the square of a very small quantity, which is much smaller,

√(1-sin(a)2) ≈ 1,

which gives an approximate equality  for small angles of the sine and tangent functions,

sin(a) ≈  tan(a).                                                     (6d)

Referring back to Fig. 2, we have

Fy (x1) = T f(t) dyn(x)/dx |x=x1,                                    (7a)

Fy (x2) = -Tf(t) dyn(x)/dx |x=x2 .                                        (7b)

The negative sign in (7b), results from the fact that the tension on the rest of the string on the left pulls leftward on the segment under consideration, which is in the negative direction, and visa versa on the right for (7a).

The net vertical force exerted on the infinitesimal element of the string is

F(x)=T f(t) [dyn(x2)-dyn(x1)].                                        (7c)

Use the relation between x1  and  x2  ,

x2= x1+ Δx,

to write

F(x)= T f(t) [dyn(x1+Δx) - dyn(x1)] ,                          (7d)

which we can rewrite as the infinitesimal

F(x)= T f(t) d2yn(x)/d2x Δx,                                           (8a)

where

x1<x<x2.

The linear mass density of the string is given by

σ =Δm/Δx.                                                             (9)

F(x)= Δm y0(x) d2f(t)/∂t2

T f(t) d2yn(x)/d2x Δx = Δm yn(x) d2f(t)/∂t2

or

T ∂2yn(x,t)/∂2x = σ ∂2yn(x,t)/∂t2

2yn(x,t)/∂2x = (σ /T) ∂2yn(x,t)/∂t2.                        (10a)

From (5b) we can differentiated twice to substitute for into both sides of (10)

2yn(x,t)/∂2t=-ωn 2 yn(x,t),

2yn(x,t)/∂2x = -(nπ/L)2yn(x,t),

to get

-(n π/L)2y(x,t) = -(σ /T) ω2 y(x,t).                                 (10b)

The frequencies of the various possible standing waves are given by (10b) solved for the angular frequency,

ωn = n π/L √(T/σ), n=1,2,3,...,∞ .                                  (11a)

Alternatively, (11a0 can be rewritten for the frequencies,

νn = n/2L √(T/σ), n=1,2,3,...,∞ .                                   (11b)

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