Suppose we want to find a function y=f(t) such that its derivatives would satisfy the equation (DE) y"+w²y=0 where a (a Greek lowercase letter called omega) is a positive constant. Such an equation involving a function y=f(t) and its derivatives together is called a differential equation. Solving a differential equation does not consist in finding a number that fits an equation, the goal is rather to find a function that fits an equation. In the above, the goal is to find which function y = f(t) satisfies equation (DE). In general, finding functions that solve a differential equation is quite hard. We sometimes offer a course on differential equations in your last term! For the specific differential equation y"+o²y=0 that we have at hand, a solution is not too difficult. As you will verify below in question 3.41B, any function of the form y(t)= acos(at)+bsin(at) (GS1) where a and b are arbitrary constants, does satisfy differential equation (DE). Note that y(t) = acos(at)+bsin(at) represents infinitely many possible functions due to the infinitely many values that the arbitrary constants a and b can take. Moreover, it is shown in more advanced mathematics courses that there are no other possible functions satisfying equation (DE), that is the infinitely many functions of equation (S1) give precisely all the possible functions that an solutions of the differential equation (DE). We call (GS1) a general solution of (DE). FODE
Suppose we want to find a function y=f(t) such that its derivatives would satisfy the equation (DE) y"+w²y=0 where a (a Greek lowercase letter called omega) is a positive constant. Such an equation involving a function y=f(t) and its derivatives together is called a differential equation. Solving a differential equation does not consist in finding a number that fits an equation, the goal is rather to find a function that fits an equation. In the above, the goal is to find which function y = f(t) satisfies equation (DE). In general, finding functions that solve a differential equation is quite hard. We sometimes offer a course on differential equations in your last term! For the specific differential equation y"+o²y=0 that we have at hand, a solution is not too difficult. As you will verify below in question 3.41B, any function of the form y(t)= acos(at)+bsin(at) (GS1) where a and b are arbitrary constants, does satisfy differential equation (DE). Note that y(t) = acos(at)+bsin(at) represents infinitely many possible functions due to the infinitely many values that the arbitrary constants a and b can take. Moreover, it is shown in more advanced mathematics courses that there are no other possible functions satisfying equation (DE), that is the infinitely many functions of equation (S1) give precisely all the possible functions that an solutions of the differential equation (DE). We call (GS1) a general solution of (DE). FODE
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Question
Can you please answer question 3.42
![Integrative Activity
Simple Harmonic Motion
This small self-learning project
is a mandatory part of the course.
Suppose we want to find a function y = f(t) such that its derivatives would satisfy the equation
y"+o²y=0
(DE)
where a (a Greek lowercase letter called omega) is a positive constant. Such an equation involving
a function y = f(t) and its derivatives together is called a differential equation. Solving a
differential equation does not consist in finding a number that fits an equation, the goal is rather to
find a function that fits an equation. In the above, the goal is to find which function y = f(t)
satisfies equation (DE).
In general, finding functions that solve a differential equation is quite hard. We sometimes offer
a course on differential equations in your last term! For the specific differential equation
y"+²y=0 that we have at hand, a solution is not too difficult. As you will verify below in
question 3.41B, any function of the form
y(t)= acos(at)+b sin(at)
(GS1)
where a and b are arbitrary constants, does satisfy differential equation (DE). Note that
y(t) = acos(at)+bsin(at) represents infinitely many possible functions due to the infinitely
many values that the arbitrary constants a and b can take. Moreover, it is shown in more advanced
mathematics courses that there are no other possible functions satisfying equation (DE), that is,
the infinitely many functions of equation (S1) give precisely all the possible functions that are
solutions of the differential equation (DE). We call (GS1) a general solution of (DE).
With some trigonometry, it can be shown that another way to write the general solution of (DE) is
(GS2)
y(t) = Acos(at + p)
where A and are constants and A is taken positive. As you will see in question 3.42 below, one
may switch between the two equivalent forms (GS1) and (GS2) of the general solution.
There are many physical situations that lead to a differential equation such as (DE), called the
(differential) equation of Simple Harmonic Motion; we will see one such example below. A
system whose position as a function of time must satisfy the equation of Simple Harmonic Motion
(DE), that is, whose position as a function of time will be given by a particular instance of equations
(GS1) or (GS2), is said to undergo Simple Harmonic Motion. Note that in (GS2) corresponds to
a "phase shift" (reason for the choice of the greak letter phi) compared to where the usual cosine
function starts:
dose
Pift
Amplitude
RTY
To aut](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2af5a58a-26e3-4baa-9073-1b18b1512b13%2F5ea62eca-bd6b-4b42-9c66-9e00ab42eebd%2Fc9x2yw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Integrative Activity
Simple Harmonic Motion
This small self-learning project
is a mandatory part of the course.
Suppose we want to find a function y = f(t) such that its derivatives would satisfy the equation
y"+o²y=0
(DE)
where a (a Greek lowercase letter called omega) is a positive constant. Such an equation involving
a function y = f(t) and its derivatives together is called a differential equation. Solving a
differential equation does not consist in finding a number that fits an equation, the goal is rather to
find a function that fits an equation. In the above, the goal is to find which function y = f(t)
satisfies equation (DE).
In general, finding functions that solve a differential equation is quite hard. We sometimes offer
a course on differential equations in your last term! For the specific differential equation
y"+²y=0 that we have at hand, a solution is not too difficult. As you will verify below in
question 3.41B, any function of the form
y(t)= acos(at)+b sin(at)
(GS1)
where a and b are arbitrary constants, does satisfy differential equation (DE). Note that
y(t) = acos(at)+bsin(at) represents infinitely many possible functions due to the infinitely
many values that the arbitrary constants a and b can take. Moreover, it is shown in more advanced
mathematics courses that there are no other possible functions satisfying equation (DE), that is,
the infinitely many functions of equation (S1) give precisely all the possible functions that are
solutions of the differential equation (DE). We call (GS1) a general solution of (DE).
With some trigonometry, it can be shown that another way to write the general solution of (DE) is
(GS2)
y(t) = Acos(at + p)
where A and are constants and A is taken positive. As you will see in question 3.42 below, one
may switch between the two equivalent forms (GS1) and (GS2) of the general solution.
There are many physical situations that lead to a differential equation such as (DE), called the
(differential) equation of Simple Harmonic Motion; we will see one such example below. A
system whose position as a function of time must satisfy the equation of Simple Harmonic Motion
(DE), that is, whose position as a function of time will be given by a particular instance of equations
(GS1) or (GS2), is said to undergo Simple Harmonic Motion. Note that in (GS2) corresponds to
a "phase shift" (reason for the choice of the greak letter phi) compared to where the usual cosine
function starts:
dose
Pift
Amplitude
RTY
To aut
![Homework Question 3.41)
A) Consider the differential equation y"+4n³y=0.
Verify by direct substitution into its left hand side that the function y=3sin(2xt) is one of its
possible solutions.
B) Similarly verify by direct substitution into the left hand side of (DE) that functions in (GS1)
do form solutions of the differential equation (DE).
C) Verify finally that functions in (GS2) similarly form solutions of (DE).
Homework Question 3.42)
The forms (GS1) and (GS2) of the general solution are actually equivalent. Indeed...
A) i) Expand the cosine in (GS2) using a trigonometric identity: the addition formula for cosine.
ii) Comparing your result in i) with equation (GS1): what are a and in terms of A and ?
B) i) Considering your expressions just obtained for a and b, add a² and b², and
use this to write A in terms of a and b.
ii) Divide the expression for b by that for a; express in terms of a and b.
Now suppose that a particle rotates at constant speed along a circle
of radius A, as shown beside.
Let T denote the period of the motion (time required to complete
a circle) and let the initial angle at time t=0 be denoted by 0.
Then the angle at any time t must be given by
0(t) = 0 +27 +
Therefore the x-coordinate of the particle is
x(t) = Acos(0) = Acos(21+0),
that is, the x-coordinate of the particle (or think of the point on the
x-axis which is the shadow projection of the point on the circle)
follows Simple Harmonic Motion with @= 2 and p=0.
Acoso
91
A is called the amplitude of the motion, its frequency is
f=+=% and is called the phase of the motion.
Thus Simple Harmonic Motion can be seen as a component of circular motion! god a
Homework Question 3.43)
Suppose the equation of motion of a particle is governed by the following differential equation of
Simple Harmonic Motion:
y"+36y=0.
A) What are the frequency and the period of the motion?
B) Write the general solution of the form (GS1) corresponding to this motion. (No work really...)
C) If at time t=0 the particle has position y(0) = 4 and velocity y'(0) = 12, find the equation
D) What is the amplitude of the motion?
of motion of the particle, that is, find the position function y(t) of the particle at any time t.
(Use the given initial conditions on position and velocity to find the appropriate a and b,
conclude by writing down the resulting formula for y(t).)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2af5a58a-26e3-4baa-9073-1b18b1512b13%2F5ea62eca-bd6b-4b42-9c66-9e00ab42eebd%2Fe2vudp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Homework Question 3.41)
A) Consider the differential equation y"+4n³y=0.
Verify by direct substitution into its left hand side that the function y=3sin(2xt) is one of its
possible solutions.
B) Similarly verify by direct substitution into the left hand side of (DE) that functions in (GS1)
do form solutions of the differential equation (DE).
C) Verify finally that functions in (GS2) similarly form solutions of (DE).
Homework Question 3.42)
The forms (GS1) and (GS2) of the general solution are actually equivalent. Indeed...
A) i) Expand the cosine in (GS2) using a trigonometric identity: the addition formula for cosine.
ii) Comparing your result in i) with equation (GS1): what are a and in terms of A and ?
B) i) Considering your expressions just obtained for a and b, add a² and b², and
use this to write A in terms of a and b.
ii) Divide the expression for b by that for a; express in terms of a and b.
Now suppose that a particle rotates at constant speed along a circle
of radius A, as shown beside.
Let T denote the period of the motion (time required to complete
a circle) and let the initial angle at time t=0 be denoted by 0.
Then the angle at any time t must be given by
0(t) = 0 +27 +
Therefore the x-coordinate of the particle is
x(t) = Acos(0) = Acos(21+0),
that is, the x-coordinate of the particle (or think of the point on the
x-axis which is the shadow projection of the point on the circle)
follows Simple Harmonic Motion with @= 2 and p=0.
Acoso
91
A is called the amplitude of the motion, its frequency is
f=+=% and is called the phase of the motion.
Thus Simple Harmonic Motion can be seen as a component of circular motion! god a
Homework Question 3.43)
Suppose the equation of motion of a particle is governed by the following differential equation of
Simple Harmonic Motion:
y"+36y=0.
A) What are the frequency and the period of the motion?
B) Write the general solution of the form (GS1) corresponding to this motion. (No work really...)
C) If at time t=0 the particle has position y(0) = 4 and velocity y'(0) = 12, find the equation
D) What is the amplitude of the motion?
of motion of the particle, that is, find the position function y(t) of the particle at any time t.
(Use the given initial conditions on position and velocity to find the appropriate a and b,
conclude by writing down the resulting formula for y(t).)
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