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Analyze the oscillations of a pendulum in a uniform gravitational field.
Solution
The picture above defines a pendulum with a thread length L and a point-mass m hanging in a gravitational field that produces a constant gravity acceleration g.
Let angle φ be a deviation of the thread from a vertical. This angle is a function of time, so we will use notation φ(t) and will attempt to include it in some equation in order to find this function.
There are two forces acting on a point-mass m: gravity, directed vertically down and equal to m·g, and tension of a thread T.
The force of gravity can be represented as a sum of two forces: directed along a thread and equal to m·g·cos(φ(t)) and perpendicular to a thread equal to m·g·sin(φ(t)).
The tension force T balances the force m·g·cos(φ(t)), as both directed along a thread in opposite directions, thus canceling each other.
The force m·g·sin(φ(t)) is the source of pendulum moving back an force, as it is always directed towards a vertical position of equilibrium.
For the purposes of this analysis we assume that initially we have moved a pendulum from its initial position, so its thread makes an angle φ(0)=φ0 with a vertical and let it go without any push, so φ'(0)=0.
According to the Second Newton's Law, the force is equal to a product of the mass and its linear acceleration.
Therefore, the force F=m·g·sin(φ(t)) that always acts along a tangential to a trajectory of the point-mass towards its equilibrium point should be equal by absolute value to a product of mass m and linear acceleration along a circular trajectory on a thread of the length L, which is equal to L·φ"(t).
Also, this force F always acts towards the vertical. Choose a positive direction of the deviation φ from a vertical as counterclockwise. For positive angles φ, as on the picture above, the direction of the force F would be negative. For negative angles φ the direction of the force F would be positive.
Therefore, the signs of the force F and deviation from the vertical are opposite to each other. For small angles φ the sign of sin(φ) would be the same as the sign of φ.
It results in the following differential equation −m·g·sin(φ(t)) = m·L·φ"(t)
or φ"(t) + (g/L)·sin(φ(t)) = 0
Since mass m can be canceled out, our first very important conclusion is that pendulum oscillation does not depend on the mass of an object hanging at its end.
The differential equation above is quite complex and we will not attempt to solve it exactly. However, as many physicists do, we can approximate it under certain conditions.
The condition usually accepted is that the angle of deviation φ(t) is very small and, consequently, function sin (φ(t)) is almost identical to function φ(t).
That brings us to an equation φ"(t) + (g/L)·φ(t) = 0
which is easy to solve, as it is a regular equation for harmonic oscillations.
The general solution to it, as explained in the previous lecture, is φ(t) = C1·cos(ωt) + C2·sin(ωt)
where ω=√g/L.
Constants C1 and C2 are determined by initial conditions.
In case of initial conditions φ(0)=φ0 and φ'(0)=0 the values of these constants are: C1 = φ0 C2 = 0
The solution of pendulum oscillations in this case is φ(t) = φ0·cos(√g/L·t)
The angular frequency of these oscillations is ω=√g/L.
The period of oscillations is T = 2π/ω = 2π√L/g.
The amplitude is A = φ0.
Just as a reminder, this analysis is applicable only to small oscillations of a pendulum around its equilibrium point, when an angle (in radians) and its sine are approximately equal to each other. Obviously, for any degree of precision the term "small" has its own meaning, the greater the precision - the smaller angle our analysis is applicable to.
Problem B
Analyze the oscillations of a ball of mass m rolling inside a semicircle of radius R in the uniform gravitational field with gravity acceleration g.
Solution
The diagram above depicts the forces acting on a ball rolling inside a semicircle.
The weight P=m·g can be represented as a sum of tangential force F=P·sin(φ(t)) and normal to a circle force that is canceled by a reaction of a circle surface.
We should take into account that the direction of force F is always opposite in sign to a sign of angle φ(t). So, the correct expressions for force F is F = −m·g·sin(φ(t))
According to Newton's Second Law, the force F can be equated to the product of mass and linear acceleration R·φ"(t) of the rolling ball. F = m·R·φ"(t)
This can be expressed as a differential equation for an angle of position φ(t). −m·g·sin(φ(t)) = m·R·φ"(t)
or, canceling mass m and dividing by radius R, φ"(t) + (g/R)·sin(φ(t)) = 0
As you see, the result is equivalent to the one we have obtained for a pendulum in the previous problem.
This differential equation is not easily solvable, so we will do exactly as in the previous problem - assume that an angle φ is so small that sin(φ(t)) can be approximated with φ(t), which leads us to a familiar differential equation for harmonic oscillations φ"(t) + (g/R)·φ(t) = 0
General and specific solutions of this equation are exactly as in the case of a pendulum in Problem A above.
Let's approach transverse waves quantitatively. More precisely, we would like to express analytically the correspondence between different characteristics of ideal transverse waves - time t, period T, angular frequency ω, speed of propagation v, amplitude A, wave length λ and wave number k.
We consider waves produced by harmonic oscillations of one end of an infinite rope, as on this picture:
(open this picture in a new tab of your browser by clicking the right button of a mouse to better see details)
For the purpose of this lecture we will simplify the real oscillations of segments of a rope and assume that each its segment is moving only transversely (perpendicularly to a stretch of a rope) and repeats the harmonic oscillation of the rope's end that drives the oscillations.
As explained in the previous lecture, the essence of waves is that any subsequent segment of a rope repeats the motion of the previous segment, but with a time delay.
The delay depends on how far segments are from each other and on the speed of propagation of the wavesv.
The speed of wave propagation, in turn, depends on many physical characteristics of a rope and is considered as given.
Let the X-axis be directed along the stretch of a rope, and X-coordinate designates the distance of a point on a rope from the end that drives the waves along a rope stretch.
Then the Y-coordinate will be a measure of a deviation of the point on a rope from the X-axis.
Assuming the initial position of the driving end of a rope is x=0 and y=0, its harmonic oscillations can be described as y(t) = A·sin(ω·t)
where (in SI units) y(t) - displacement (meters) A - amplitude (meters) ω - angular frequency (rad/sec) t - time (sec)
For any other point on a rope the displacement (deviation from X-axis, oscillations, wave function) y() depends not only on time t, but also on a distance x from the rope's end that drives the oscillations, because the time delay of this point's motion relatively to motions of the rope's end depends on this distance. So, it's appropriate to analyze the function y(x,t) - a displacement of a point of a rope on a distance x from the rope's driving end at moment of time t.
In terms of the wave function y(x,t), the oscillations of the rope's end (x=0) are described as y(0,t) = A·sin(ω·t)
The parameters that must be given to define the oscillations y(x,t) for any x and t are: A - amplitude ω - angular frequency and v - wave propagation speed
The first two determine the oscillations of the driving end of a rope.
The wave propagation speed will help to define the oscillations of any other point on a rope.
Using these parameters, we can determine the wave function y(x,t) of any point of a rope.
As mentioned before, every point of a rope repeats the motions of the rope's driving end with a delay that depends on a distance between this point and the driving end of a rope x.
If the speed of wave propagation is v, the time delay will be x/v.
So, function y(x,t) should be equal to y(0,t−x/v).
Indeed, assume that at t=t0 the driving end of a rope is at Y-coordinatey(0,t0).
To have that same value at distance x from the rope's end at the time moment t0+x/v we have to set function y(x,t) to y(x,t0) = y(0,t0−x/v)
Then y(x,t0+x/v) = y(0,t−x/v+x/v) =
= y(0,t0)
which is exactly what we need - at t=t0+x/v - function y(x,t) equals to y(0,t0).
So, our final expression for a wave function that represents the Y-coordinate of a point at distance x from the rope's end at time t is y(x,t) = y(0,t−x/v) =
= A·sin(ω·(t−x/v))
This form of wave equation can be transformed in many different ways, using different parameters and expressions of one parameter in terms of others.
Recall the following definitions.
The wave length λ is the distance between two consecutive wave crests or troughs.
The period T is the time a wave propagates by a distance equal to its wave length λ moving with its wave propagation speed v, that is T=λ/v.
On the other hand, if we consider a function sin(ω·t), it has a period ω times smaller than function sin(t). The latter has a period 2π. Therefore, function sin(ω·t) has a period T=2π/ω and, therefore, ω=2π/T=2π·v/λ.
The above formulas allow us to transform the obtained wave equation y(x,t)=A·sin(ω·(t−x/v))
into y(x,t) = A·sin(2π·(t−x/v)/T) =
= A·sin(2π·(t−x/v)·v/λ) =
= A·sin(2π·(v·t−x)/λ)
The quantity k=2π/λ is called wave number. From this definition follows that k·v = (2π/λ)·(λ/T) = 2π/T = ω
This can be used in yet another form of the wave equation y(x,t) = A·sin(k·(v·t−x)) =
= A·sin(ω·t−k·x))
From this we conclude that a point at distance x from the driving end of a rope performs the same oscillating motions as the rope's driving end, but shifted by phase equal to k·x, where k is a wave number.
All the above forms of the wave equation are equivalent to each other and are used interchangeably.
We have started with harmonic oscillations of a driving end of a rope as y(0,t)=A·sin(ω·t). It was based on initial position of the rope's end y(0,0) at Y-coordinate equal to zero and initial speed (first derivative by time) equal to y'(0,0)=A·ω·cos(ω·0)=A·ω
(not zero).
Another possibility would be to start from y(0,0)=A and no initial speed y'(0,0)=0.
To satisfy these initial conditions of the harmonic oscillations we can choose a function y(0,t)=A·cos(ω·t).
Then the wave equation takes a different form y(x,t) = y(0,t−x/v) =
= A·cos(ω·(t−x/v)) =
= A·cos(2π·(t−x/v)/T) =
= A·cos(2π·(t−x/v)·v/λ) =
= A·cos(2π·(v·t−x)/λ) =
= A·cos(k·(v·t−x)) =
= A·cos(ω·t−k·x))
As you see, the general sinusoidal behavior is preserved in this case, as sine and cosine functions behave pretty much the same way with only difference in phase shift.
Finally, let us point again that the above analysis is based on a simplified model of transverse oscillations with every point along a rope performing harmonic oscillation, repeating the harmonic oscillations of a driving end of a rope with some time delay (or phase shift) that depends on wave propagation speed.
Reality, as usually, is much more complex than our models.
In the previous lecture we considered the model of a musical string as a point-mass m between two identical springs of the length (L+l)/2 each (where L is a neutral length of the string we are modeling and l is a stretch to create initial tension) with a coefficient of elasticity k.
Plucking a string is, therefore, modeled as an initial displacement of a point-mass vertically by y(0)=a.
The result of the analysis of forces acting on a point-mass between these springs, when it performs transverse oscillations, was a differential equation y"(t) + 2·(κ/m)·y(t)·
·[1−L/√(L+l)²+4y²(t)] = 0
Let's examine this model from the energy standpoint.
By lifting a point-mass initially by distance y(0)=a, we perform some work against elastic forces of two springs. The forces are variable and depend on how much we stretch the springs. Let's take coordinate y as our argument. Then, as explained in the previous lecture, the forces of two springs acting along the Y-axis against our motion total to F↓↓(y) = −2·κ·y·
·[1 − (L/2) /√((L+l)/2)²+y²] =
= −2·κ·y·[1 − L /√(L+l)²+4y²]
Incidentally, it's easy to see that an expression in [...] is always positive, therefore the sign of the force is always opposite to the sign of a displacement y.
On an interval from y to y+dy of the infinitesimal length dy we can assume the constant force, as described above.
Therefore, the work we perform on that infinitesimal interval by going against the force F↓↓(y) is dW = −F↓↓(y)·dy
We can now calculate the total work performed by displacing our point-mass by distance y(0)=a by integrating the above expression from y=0 to y=a. W[0,a] = −∫[0,a]F↓↓(y)·dy =
= ∫[0,a]2·κ·y·
·[1−(L/2) /√((L+l)/2)²+y² ]·dy =
= ∫[0,a]2·κ·y·
·[1−L/√(L+l)²+4y² ]·dy
This integral can be calculated.
The answer is W[0,a] = 2·κ·[y²/2 −
− (L/4)·√(L+l)²+4y²] |[0,a] =
= 2·κ·[a²/2 −
− (L/4)·√(L+l)²+4a² +
+ L·(L+l)/4] =
= κ·[a² − (L/2)·√(L+l)²+4a² +
+ L·(L+l)/2]
Just to check, if a=0, the work performed is W[0,0] =
= κ·[0² − (L/2)·√(L+l)²+0² +
+ L·(L+l)/2] = 0
as it is supposed to be.
Analysis of the potential energy W(a) as a function of the initial displacement a shows that it resembles the parabola with a graph like this:
So, potential energy is always positive, as it should be, and grows with initial displacement a growing to positive or negative direction (up or down on a picture of a point-mass between two springs above).
At any displacement point y from the neutral position a point-mass m has certain potential and kinetic energy.
Its potential energy W(y), as a function of displacement y, looks exactly the same as the one we calculated at the initial position, when y=a, which yields W(y) =
= κ·[y² − (L/2)·√(L+l)²+4y² +
+ L·(L+l)/2]
Obviously, this formula is valid only for −a ≤ y ≤ a.
The difference between W[0,a] and W(y) is the kinetic energy our point-mass possesses because it moves towards the neutral position, that is K(y)=m·(y')²/2.
Therefore, we have an equation, which is a differential equation for a displacement y(t) as a function of time t (in many cases we will drop (t) for brevity). W(y) + K(y) = W[0,a]
The differential equation for displacement y(t) is κ·[y² − (L/2)·√L+l)²+4y² +
+ L·(L+l)/2] + m·(y')²/2 =
= κ·[a² − (L/2)·√(L+l)²+4a² +
+ L·(L+l)/2]
This is a different differential equation that describes the transverse oscillations of a point-mass between two springs than that described in the previous lecture.
It contains only the first derivative of a displacement y(t).
Interestingly, though not surprisingly, this equation is not really different.
If two functions are equal, their derivatives are also equal. Take a derivative from both sides of an equation above: (m/2κ)·2y'·y" + 2y·y' −
− (L/2)·8y·y' /2√(L+l)² + 4y² =
= 0
or (m/2κ)·y'·y" + y·y' −
− L·y·y' /√(L+l)² + 4y² =
= 0
Now we can cancel y'(t) getting (m/2κ)·y" + y −
− L·y /√(L+l)² + 4y² = 0
or (m/2κ)·y" + y·
·[1 − L /√(L+l)² + 4y²] = 0
or y" + (2κ/m)·y·
·[1 − L /√(L+l)² + 4y²] = 0
which is exactly the same as the equation derived in the previous lecture.
Our first example of transverse oscillations was the one with a rope, one end of which we shake up and down, with another end free.
(open this picture in a new tab of your browser by clicking the right button of a mouse to better see details)
Now we will consider another type of transverse oscillations that is used in string musical instruments, like a guitar or violin.
The real oscillations of a violin string are quite complex, so we will have to rely on some simplified model that allows analytical approach.
So, here is the first really simple model of an oscillating string.
String of a violin or a guitar has two ends fixed and it oscillates when it's plucked.
The mechanism of plucking involves stretching a string at some middle point perpendicularly to a string direction and letting it go.
It is important to notice that real guitar or violin strings are initially stretched to create some tension, without which there will be no pleasant sound.
Also important is that real strings have a high coefficient of elasticity, so even a small stretch creates significant tension force.
In our simplified model we replace a string with two identical weightless springs of combined length L in a neutral (not stretched, nor squeezed) state, but stretched by a combined increment l to create initial tension, and a point-mass m fixed in the middle between the springs.
So, the neutral length of each spring is L/2 and the length increment of each spring, as it is initially stretched, is l/2.
According to the Hooke's Law, the tension caused by initial stretch of each spring is T = k·l/2
The initial tension forces of both springs act on a point-mass in the middle with equal magnitude and opposite directions and neutralize each other.
Plucking the point-mass in the upward direction, thus further stretching both springs it's attached to, and letting it go resembles the plucking of a string on a musical instrument, like a guitar or a violin.
The point-mass will start oscillating in the vertical direction, performing transversal oscillations.
This simplified model allows for analytical approach to find the differential equation of this type of oscillation.
Let's state in advance that the differential equation obtained will not yield to an easy solution, so we will just explain how to get to it, but then switch to another way to approach this problem.
Let's examine the forces acting on our point-mass after we lift it by the initial distance a and let it go, so it's performing vertical oscillations with y(t) being a deviation from the initial position at time t.
The initial condition of this motion is, therefore, y(0) = a y'(0) = 0
Assume that at time t the vertical deviation of a point-mass from its initial position is y(t). Then the two forces from two stretched springs acting on this point-mass are in red on a picture below.
The forces can be evaluated using the Hooke's Law.
The neutral (unstretched) length of each spring is L/2, but it's initially stretched by l/2. The new stretched length is √((L+l)/2)²+y²(t). The force is proportional to elongation of a spring with a coefficient κ - spring's elasticity.
Therefore, each force by absolute value equals to |F(t)| =
= κ·[√((L+l)/2)²+y²(t) − L/2]
We need to account only for vertical components of these two forces since horizontal components will act against each other and neutralize each other.
If each spring makes an angle θ with horizontal line, the absolute value of the vertical component of each force is |F↓(t)| = |F(t)|·sin(θ)
In terms of y(t) the vertical component of both forces is F↓(t) =
= −|F(t)|·y(t) /√((L+l)/2)²+y²(t)
Since vertical components from both springs are added together, the total force F↓↓(t) acting on a point-mass is double the above expression, which should be simplified as follows. F↓↓(t) = m·y"(t) =
= −2·|F(t)|·
·y(t) /√(L+l/2)²+y²(t) =
= −2·κ·y(t)·
·[1 − (L/2) /√(L+l/2)²+y²(t)]
This gives us a differential equation of a transverse movement of a point-mass between two springs m·y"(t) = −2·κ·y(t)·
·[1 − (L/2) /√((L+l)/2)²+y²(t)]
In a more traditional form y"(t) + 2·(κ/m)·y(t)·
·[1−(L/2)/√((L+l)/2)²+y²(t)] = 0
If we consider the above equation as it is, it's too complex to get a nice solution. In cases like this physicists usually resort to reasonable approximations.
In this case we can safely assume that vertical deviation y(t) is small relatively to a the length L and, therefore, adds an insignificant amount to a radical in the denominator. So, let's just drop it from the equation, getting after a series of trivial steps a simpler differential equation
(a) y"(t) + 2·(κ/m)·y(t)·
·[1−(L/2)/√((L+l)/2)²] = 0
(b) y"(t) + 2·(κ/m)·y(t)·
·[1−(L/2)/((L+l)/2)] = 0
(c) y"(t) + 2·(κ/m)·y(t)·
·[1−L/(L+l)] = 0
(d) y"(t) + 2·(κ/m)·y(t)·
·l /(L+l) = 0
(e) y"(t) + ω²·y(t) = 0
where ω² = 2·κ/[m·(1+L/l)]
The final equation is a familiar differential equation that describes harmonic oscillations with angular frequency ω.
From the expression ω² = 2·κ/[m·(1+L/l)]
we see that greater initial stretch of our springs l contributes to higher angular frequency of oscillations, which corresponds to our experience with real strings - more initial tension applied to a string results in a higher tone of vibrations.
Considering initial conditions y(0) = a and y'(0) = 0,
the vertical displacement of our point-mass between two springs will be y(t) = a·cos(ωt)
where ω is evaluated above.
We are familiar with longitudinal waves, like sound waves in the air.
Their defining characteristic is that molecules of air (the medium) are oscillating along the direction of the wave propagation, which, in turn, causing oscillation of pressure at any point along the direction of wave propagation.
Consider a different type of waves.
Take a long rope by one end, stretching its length on the floor. Make a quick up and down movement of the rope's end that you hold.
The result will be a wave propagating along the rope, but the elements of rope will move up and down, perpendicularly to the direction of waves propagation.
(open this picture in a new tab of your browser by clicking the right button of a mouse to better see details)
These waves, when the elements of medium (a rope in our example) are moving perpendicularly to a direction of waves propagation, are called transverse.
Such elementary characteristics of transverse waves as crest, trough, wavelength and amplitude are clearly defined on the picture above.
Some other examples of transverse waves are strings of violin or any other string musical instrument.
Interesting waves are those on the surface of water. They seem to be transverse, but, actually, the movement of water molecules is more complex and constitutes an elliptical kind of motion in two directions - up and down perpendicularly to waves propagation and back and forth along this direction.
Our first problem in analyzing transverse waves is to come up with a model that resembles the real thing (like waves on a rope), but yielding to some analytical approach.
Let's model a rope as a set of very small elements that have certain mass and connected by very short weightless links - sort of a long necklace of beads.
Every bead on this necklace is a point-object of mass m, every link between beads is a solid weightless rod of length r.
Both mass of an individual bead m and length of each link r are, presumably, very small. In theory, it would be appropriate to assume them to be infinitesimally small.
Even this model is too complex to analyze. Let's start with a simpler case of only two beads linked by a solid weightless rod.
Our purpose is not to present a complete analytical picture of waves, using this model, but to demonstrate that waves exist and that they propagate.
Consider the following details of this model.
Two identical point-objects α and β of mass m each on the coordinate plane with no friction are connected with a solid weightless rod of length r.
Let's assume that α, initially, is at coordinates (0,−A), where A is some positive number and β is at coordinates (a, −A).
So, both are at level y=−A, separated by a horizontal rod of length r from x=0 for α to x=a for β.
We will analyze what happens if we move the point-object α up and down along the vertical Y-axis (perpendicular to X-axis), according to some periodic oscillations, like yα(t) = −A·cos(ω·t)
where A is an amplitude of oscillations, ω is angular frequency, t is time.
Incidentally, for an angular frequency ω the period of oscillation is T=2π/ω.
So, the oscillations of point α can be described as yα(t) = −A·cos(2π·t / T)
Point α in this model moves along the Y-axis between y=−A and y=A.
It's speed is y'α(t) = A·ω·sin(ω·t) =
= A·(2π/T)·sin(2π·t/T)
The period of oscillations is, as we noted above T = 2π/ω
During the first quarter of a period α moves up, increasing its speed from v=0 at y=−A to v=A·ω at the point y=0, then during the next quarter of a period it continues going up, but its speed will decrease from v=A·ω to v=0 at the top most point y=A.
During the third quarter of a period α moves down from y=A, increasing absolute value of its speed in the negative direction of the Y-axis from v=0 to the same v=A·ω at y=0, then during the fourth quarter of a period it continues going down, but the absolute value of its speed will decrease from v=A·ω to v=0 at the bottom y=−A.
Positions of objects α and β during the first quarter of a period of motion of α at three consecutive moments in time are presented below.
As the leading object α starts moving up along Y-axis, the led by it object β follows it, as seen on a picture above.
Let's analyze the forces acting on each object in this model.
Object β is moved by two forces: tension from the solid rod Tβ(t), directed along the rod towards variable position of α, and constant weight Pβ.
Object α experiences the tension force Tα(t), which is exact opposite to Tβ(t), the constant weight Pα and pulling force Fα(t) that moves an entire system up and down.
There is a very important detail that can be inferred from analyzing these forces.
The tension force Tα(t), acting on object α, has vertical and horizontal components from which follows that pulling force Fα(t) cannot be strictly vertical to move object α along the Y-axis, it must have a horizontal component to neutralize the horizontal component of Tα(t).
This can be achieved by having some railing along the Y-axis that prevents α to deviate from the vertical path. The reaction of railing will always neutralize the horizontal component of the Tα(t). Without this railing the pulling force Fα(t) must have a horizontal component to keep α on the vertical path along the Y-axis.
If α and β are the first and the second beads on a necklace, we can arrange the railing for α. But, if we continue our model and analyze the movement of the third bead γ attached to β, there can be no railing and the horizontal component of the tension force Tβ(t) will exist and will get involved on some small scale.
All-in-all, transverse motion is not just movement of components up and down perpendicularly to the wave propagation, it is also a longitudinal motion of these components, though not very significant in comparison with transverse motion and often ignored in textbooks.
The really obvious reason for transverse motion to involve a minor horizontal movement in addition to a major vertical one is that you cannot lift up a part of a horizontally stretched rope without a little horizontal shift of its parts as well, because a straight line is always shorter than a curve.
Let's now follow the motion of β as α periodically moves up and down, starting at point (0,−A), according to a formula yα(t) = −A·cos(ω·t)
with its X-coordinate always being equal to zero.
Since position of α is specified as a function of time and the length of a rod connecting α and β is fixed and equal to r, position of β can be expressed in term of a single variable - the angle φ(t) from a vector parallel to a rod directed from α to β and the positive direction of the Y-axis.
At initial position, when the rod is horizontal, φ(0)=π/2.
Then coordinates of β are: xβ(t) = r·sin(φ(t)) yβ(t) = yα(t) + r·cos(φ(t))
During the first quarter of a period α moves up from y=−A to y=0, gradually increasing its speed and pulling β by the rod upwards.
While β follows α up, it also moves closer to the Y-axis. The reason for this is that the only forces acting on β are the tension force along the rod and its weight.
Weight is vertical force, while tension acts along the rod and it has vertical (up) and horizontal (left) components. Assuming vertical pull is sufficient to overcome the weight, β will be pulled up and to the left, closer to the Y-axis.
If this process of constant acceleration of α continued indefinitely, β would be pulled up and asymptotically close to the Y-axis. Eventually, it will just follow α along almost the same vertical trajectory upward.
The angle φ(t) in this process would gradually approach π (180°).
With a periodic movement of α up and down the Y-axis, the trajectory of β is much more complex.
Let's analyze the second quarter of the period of α's oscillations, when it moves from y=0 level, crossing the X-axis, to y=A.
After α crosses the X-axis it starts to slow down, decelerate, while still moving up the Y-axis.
As α decelerates in its vertical motion up, the composition of forces changes.
Now β will continue following α upwards, but, instead of being pulled by the rod, it will push the rod since α slows down.
That will result in the force Tα(t), with which a rod pushes α, to be directed upwards and a little to the left, as presented on the above picture.
Also Tβ(t), the reaction of the rod onto β, will act opposite to a tension during the first quarter of a period. Now it has a vertical component down to decelerate β's upward motion and a horizontal component to the right.
The latter will cause β to start moving away from the Y-axis, while still going up for some time, slowing this upward movement because of opposite force of reaction of the rod.
When α reaches level y=A at the end of the second quarter of its period, it momentarily stops.
The behavior of β at that time depends on many factors - its mass m, amplitude A and angular frequency ω of α's oscillations, the length of the rod r.
Fast moving leading object α will usually result in a longer trajectory for β, while slow α will cause β to stop sooner.
During the third quarter of its period α starts moving downward, which will cause β to follow it, but not immediately because of inertia. Depending on factors described above, β might go up even higher than α. In any case, the delay in β's reaching its maximum after α has completed the second quarter of a period will exist, that would cause β to reach its maximum later than α.
The delay between reaching its maximum height by β is the source of a visual effect that call waves. Every subsequent bead on a necklace will reach its top height a little later than the previous.
Assuming that α accelerates faster than g on its way down during the third quarter of a period, α will pull β down and closer to the Y-axis. Situation will be similar to the first quarter of a period, when α accelerates up.
Then during the fourth quarter α will slow down to full stop, while inertia and weight move β faster down, so it will have to make a small circle around the last level of α before a new period starts, and α pulls β again up.
The above relatively complicated description of the trajectory of β is supposed to serve as an explanation of the fact that this trajectory represents not simply an up and down motion of beads of a necklace in our model of the transverse oscillations, but is much more complex motion that involves both transverse and longitudinal oscillations with former playing a major role, while latter, however small, is still an important part of a motion.
Let's consider the same simple device as in the previous lecture - a cylinder of relatively small diameter with air and a piston in it, except this time we will examine what happens at the opposite to a piston end of a cylinder.
Let's give a single short and quick push to a piston down. As a result, a single parallel to a piston wave of a higher pressure air will start its way down a cylinder with certain speed v that depends on properties of air, its temperature, initial pressure etc.
When this wave reaches the bottom of a cylinder, a thin layer of higher pressure air will develop there. Since the bottom of a cylinder prevents a wave to go farther, the higher pressure air will release its pressure back to a layer above it. This constitutes a reflection of the wave.
After such a reflection the wave will go to the opposite direction (upward on the picture) until it reaches the top, where piston is positioned. At this moment, since piston is not moving, it will play the same role as a bottom of a cylinder before, it will reflect the wave.
Ideally, this reciprocal movement of a wave can continue, it will be reflected from both ends of a cylinder indefinitely.
Practically, the energy of the initial push of a piston will gradually dissipate into more intense movement of the air molecules, which means that the temperature of the air will increase slightly.
Consider now a harmonic oscillation of a piston. It produces pressure waves in the air propagated downward. But now we know that they reflect from the bottom of a cylinder and propagate upwards.
This results in a very complex superposition of two sets of waves, one going downwards and another upwards.
There is, generally speaking, a shift in phase between these two sets of waves that depends on the speed of wave propagation and the length of a cylinder.
The resulting picture of air pressure will look quite chaotic and far from harmonic oscillations of a piston. Yet, it's just a simple superposition of waves moving in two opposite directions.
Imagine a complexity of a sound distribution in an opera house with air pressure waves, originated by a person singing on a stage, reflected from all hard surfaces and mixing in time and space with other waves. Unless special design measures are taken, the quality of a sound will be very low.
Even in such a simple case as recording these lectures we had to use special directional microphone that suppresses sounds from all directions except the main one directed onto a primary source of sound. In addition, to weaken the reflection, we open closets on the opposite wall to weaken the strength of reflected sound waves.
Let's analyze the reflection of a simple flat wave from a surface that is not perpendicular to the wave front, keeping in mind the Law of Reflection, that is, the angle of incidence equals to the angle of reflection.
Assume that our flat wave consists of only two molecules of air a and b that hit a surface AB at 45° angle.
A thin line between these molecules represents a wave's front, which is perpendicular to their trajectories.
From a picture above and some elementary geometry it's obvious that reflected wave will be exactly as original, except its direction will be perpendicular to the original's one.
Indeed, a and b move on parallel trajectories with the same speed, covering the same distance in equal times. So, molecule a will come to point C at the same time as molecule b reaches point B. After that a and b continue their parallel movement and the new wave's front is still perpendicular to molecules' trajectories.
Even if the angle of a reflecting surface is not 45° to a direction of the original wave, the reflected wave will still be a flat wave with only direction changed.
It's easy to prove that molecule a will reach point C at the same time as molecule b comes to point B. After that both molecules continue their straight movement with a wave front perpendicular to their parallel trajectories.
We should not think about waves and their reflections as a flow of the same molecules toward a reflective surface, reflecting from it and continuing the movement. Different crests of waves have different molecules. But the movement of one molecule is transferred to its neighbor in the direction of a wave propagation as if they are connected with tiny springs.
Another good comparison of this process is the balls on a billiard table. When a ball hits another ball straight on, it stops and the other ball continues the movement in the same direction. If the hit is not exactly straight on, the impulse of a hitting ball will be distributed among two balls, but their sum will be equal to the original impulse.
So, the energy and impulse are transferred in some direction without carriers of this energy and impulse actually moving all the way to the end of a road, but rather by transferring this energy and impulse to other carriers, like in the relay race.
Let's return back to a case when a flat solid surface is perpendicular to a direction of propagation of the flat waves, like a bottom of a cylinder described above.
As mentioned above, reflected from this obstacle waves will be reflected in the opposite direction, thus interfering with original waves.
Both oscillations, downwards and upwards, have exactly the same period T and wavelength λ. Their speeds v are equal by absolute value, but oppositely directed.
Depending on difference in phase, superposition of these two waves can produce different interesting effects.
An oscillation of air in a cylinder, produced by an oscillation of a piston, is just a periodic higher and lower concentration of molecules of air (that is, higher or lower pressure) with the waves of high concentration moving from a piston down a cylinder with some speed.
Add to this picture reflected waves formed at the opposite end of a cylinder - waves of higher and lower concentration of molecules of air (that is, higher or lower pressure) moving up the cylinder.
Both processes mix together in a superposition. When a wave crest from a piston reaches some layer of air, it pushes this layer down.
When the opposite reflected wave crest reaches the same layer, it pushes this layer up.
With proper difference in phase, that depends on the length of cylinder relatively to a wavelength λ, there might be a situation when the same layer of air in a cylinder is simultaneously pushed down by a wave crest from a piston and up by a reflected wave crest. As a result, this layer will be squeezed from both ends, the pressure in this layer will become higher.
For similar reasons the layer on a distance λ/2 (half a wavelength) from the one that is squeezed from both sides will be thinner, and the pressure inside this layer will be lower.
In T/2 (half a period) time the area that used to be with higher concentration of molecules will become thinner, as its molecules will move in opposite directions, thus making areas on a distance of half a wavelength around become more concentrated, that is with higher pressure.
This is a case of waves formed by two opposite sources of oscillation - by a piston and by a reflection from the bottom of a cylinder.
The distance between a piston and a bottom of a cylinder will be divided in layers of half a wavelength thick. Each layer will be squeezed and stretched every half a period time, becoming more concentrated and less concentrated than its neighbors.
Each layer would look like an accordion gaining some air inside and squeezing it out every half a period time.
So, entire air in a cylinder can be viewed as a series of sidewise connected accordions. When every other one is synchronously squeezed, both its neighbors are stretched and, when every other one is synchronously stretched, its both neighbors squeezed.
The crests of these waves of higher and lower pressure will not travel, as in a case without reflection, but will stand on the same place. This phenomenon is known as standing waves.
Pressure waves are longitudinal oscillations of molecules of some medium caused by mechanical vibration of a source and propagated through medium.
Let's consider a simple device - a long cylinder of relatively small diameter with air and a piston in it.
Initially, a piston is in the top position, and air molecules are chaotically moving inside, bumping into each other, hitting the walls of a cylinder and a piston, thus creating certain pressure on them.
Their "chaotic" behavior just means that they are moving in all directions with equal probabilities.
The pressure inside a cylinder depends on how much air is in it, the volume of a cylinder and its temperature. Assume for simplicity that amount of air and its temperature are constant.
Then in a short motion we quickly push a piston, abruptly reducing the volume of a cylinder.
If we did it slowly, the molecules of air would have the time to relatively evenly redistribute inside a smaller volume of a cylinder. This, according to the Boyle's Law, would cause the pressure inside a cylinder to evenly increase at every point inside.
If, however, we did it in a short quick motion, the pressure immediately under a piston will rise, while farther from a piston will not, simply because air molecules in immediate vicinity of a piston will not have time to evenly redistribute throughout the whole cylinder's new (smaller) volume.
Statistically, the probability of moving in a direction away from a moving piston will be higher for those molecules immediately under it, which will create extra pressure on their neighbors down the cylinder.
After transferring part of their kinetic energy to neighboring molecules down a cylinder, those molecules immediately under a piston will gradually restore their chaotic movement to all directions with equal probabilities, but their neighbors, receiving extra push from above, will have their chaotic movement changed in a way that they will move down a cylinder with greater probability than up.
This process of transferring kinetic energy continues down a cylinder, from one layer of air molecules to another, thus creating a flat wave of area of higher pressure.
This is analogous to easily observed process of hooking a locomotive to a train with many cars. Cars are connected with a slight gap in a link between them, so locomotive, when hooked to the first car, hits and moves it a little. Because of a gap in a link mechanism the energy of a locomotive is transferred from the first car to the next with a slight delay needed for the first car to close the gap with the second. Then another delay between the second and third etc.
The speed of this wave moving down a cylinder depends on many factors, mostly it depends on density of air molecules inside a cylinder and, obviously, certain characteristics of molecules themselves and how they interact. To a lesser extent it depends on a speed of moving piston.
When a piston stops, the air molecules immediately under it will continue moving down a cylinder for awhile, which creates an area of lower pressure immediately under a piston. The higher pressure from the next layer of molecules will push molecules under a piston back to their original place, which, in turn, creates lower pressure in the next layer of molecules down a cylinder.
So, lower pressure, formed under a piston immediately after it stopped its downward motion will travel down the cylinder. It will create a new flat wave of lower pressure that follows the high.
Now consider that we quickly moved a piston up. During the movement of a piston up the cylinder the air pressure immediately under it will be smaller than further down a cylinder, because air molecules need some time to fill newly increased volume of space.
Statistically, the probability of moving in a direction towards a moving piston will be higher for those molecules immediately under it, so they will start moving upwards, following the cylinder movement, as if a piston pulled them up.
They will fill the area of a lower pressure under the piston, thereby creating a lower pressure below, so the area of lower pressure will move down a cylinder as a wave.
When a piston stops, the air molecules immediately under it will continue moving up a cylinder for awhile, which creates an area of higher pressure immediately under a piston. This higher pressure will start pushing the next layer of molecules under it down, which, in turn, creates higher pressure in the next layer of molecules.
So, higher pressure, formed under a piston immediately after it stopped its upward motion will travel down the cylinder. It will create a new flat wave of higher pressure that follows the low pressure wave.
Finally, consider quick reciprocal movement of a piston back and forth. If this movement is faster than normal time for air molecules to evenly distribute within the available volume, zones of higher and lower pressure will be produced by a piston and will propagate towards an opposite side of a cylinder. These are pressure waves.
On the picture above different areas of air pressure are presented in different colors from high pressure in blue to low pressure in red.
Let's introduce some wave characteristics, assuming that the piston in our experiment is periodically and reciprocally moving back and forth significantly faster than air molecules inside a cylinder can evenly distribute.
We also assume that a cylinder is long enough (strictly speaking, of infinite length, that is, bottomless), so we don't have to worry about a reflection wave from its bottom that superimposes on the primary wave initiated by a piston.
These are conditions when waves of high and lower pressure are following each other with constant frequency and with constant distance between the peaks.
The first parameter that we introduce to characterize the waves is the wave length, which is a distance from one peak of pressure to the next one. We can talk about distance as the length between the peaks, that is wave length λ, and the time that passes between one wave crest passes a particular point and the next one, which is wave period T .
From these two parameters of waves we derive wave speed v=λ/T and frequency ν=1/T.
Let's represent graphically the waves in a medium in terms of pressure p as a function of time t and distance d from a piston, the source of oscillation.
In a cylinder with a piston the pressure of air is inversely proportional to the volume of a part of a cylinder under a piston, which, in turn, is proportional to a position of a piston, since cross section of a cylinder is constant.
Assume, the piston is reciprocally moving in harmonic oscillations, and its very small displacement Δy(t) relatively to some initial position can be described as a function of time t as
Δy(t) = A·sin(ω·t)
where A is a very small amplitude of deviations from initial position and ω=2π/T is an angular frequency of piston's oscillations.
We have chosen to use a function sin, so at time t=0 the deviation from initial position is zero.
Let's limit the oscillations to only one period (one back and forth movement of a piston) and assume that the time of a period T is very small, so we can examine the behavior of a single wave.
Let's examine the air in the area immediately under a piston - a very thin layer of molecules having initial volume V0.
We can assume that within this very thin layer the pressure of air during piston's movement is the same and only depends on its volume, which, in turn, is a function of the position of a piston. Farther down a cylinder the situation will be different since areas of high and low pressure will not travel immediately, but with some speed that depends on the property of air.
Obviously, a change of volume of a thin layer of air immediately under a piston ΔV(t) can also be represented as harmonic oscillations around initial value V0
ΔV = B·sin(ω·t)
where B is a very small amplitude of deviations from initial volume and ω is the same as above angular frequency of piston's oscillations.
So, the volume of our thin layer of air, as a function of time, is V(t) = V0 + B·sin(ω·t)
where V0 is the initial volume of our thin layer of air at time t=0.
According to Boyle's Law, air pressure p(t) in the thin layer under a piston that we consider is inversely proportional to its volume.
Therefore, p(t) = C / V(t)
where C is some constant.
From this follows p(t) = C / V(t) =
= C / [V0 + B·sin(ω·t)]
The expression above might not look like harmonic oscillation, but, actually, it's very close to it.
Firstly, it's periodic because the numerator is a constant and the denominator is a periodic function.
Secondly, it oscillates between its maximum, when the denominator at minimum (when sin(ω·t)=−1) pmax = C / [V0 − B]
to its minimum, when the denominator at maximum (when sin(ω·t)=+1) pmin = C / [V0 + B]
So, we do obtain oscillations of the pressure in the area immediately adjacent to a piston, just not exactly harmonic oscillations.
Knowing the pressure in the layer immediately adjacent to a piston as a function of time p(t) and the speed v of waves propagation (that depends on the amount and characteristics of air in the cylinder), we can come up with a wave function on a distance x from an oscillating piston as a function P(x,t) of both distance x and time t.
The distance x from the piston introduces a time delay τ=x/v to a wave function, so pressure at distance x will be characterized by the same function p(t) with a shift in time by τ. P(x,t) = p(t+τ) = p(t+x/v) =
= C / [V0 + B·sin(ω·(t+x/v))]
Graphically, this function P(x,t) looks like this
As we see, the same pressure wave goes through points on any distance from the initial position of a piston, but comes there later in time.
And, the farther the point is from a piston - the later the pressure wave comes to it.
