## Saturday, October 30, 2021

### Problems 1: UNIZOR.COM - Physics4Teens - Waves - Mechanical Oscillations

Notes to a video lecture on http://www.unizor.com

Problems 1

Problem A

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.

## Saturday, October 23, 2021

### Wave Equation 2: UNIZOR.COM - Physics4Teens - Waves - Transverse Waves

Notes to a video lecture on http://www.unizor.com

Wave Equation 2

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 waves v.
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)
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-coordinate y(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 . 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.

## Sunday, October 17, 2021

### Musical Strings 2: UNIZOR.COM - Physics4Teens - Waves - Transversal Waves

Notes to a video lecture on http://www.unizor.com

Musical Strings 2

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.
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
]

Simplifying...
y² − (L/2)·√(L+l)²+4y² +
+ (m/κ)·(y')²/2 =
= a² − (L/2)·√(L+l)²+4a²

or
(m/2κ)·(y')² + y² −
− (L/2)·√(L+l)² + 4y² =
= a² − (L/2)·√(L+l)²+4a²

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.

## Friday, October 8, 2021

### Musical Strings 1: UNIZOR.COM - Physics4Teens - Waves - Transversal Waves

Notes to a video lecture on http://www.unizor.com

Musical Strings 1

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)]

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.