Show that the x component of the particle's position executes simple harmonic motion
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solving some excercise from Richard Haberman Mathematical Models, but I don't seems to know what to do to this problem.
mathematical-physics classical-mechanics
asked Jan 22 at 19:33
Niang MooreNiang Moore
326
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add a comment |
$begingroup$
solving some excercise from Richard Haberman Mathematical Models, but I don't seems to know what to do to this problem.
mathematical-physics classical-mechanics
asked Jan 22 at 19:33
Niang MooreNiang Moore
326
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Draw a picture and break position into components.
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– David G. Stork
Jan 22 at 19:37
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What is the definition of simple harmonic motion, in that text?
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– Aretino
Jan 22 at 20:14
add a comment |
$begingroup$
solving some excercise from Richard Haberman Mathematical Models, but I don't seems to know what to do to this problem.
mathematical-physics classical-mechanics
asked Jan 22 at 19:33
Niang MooreNiang Moore
326
$endgroup$
solving some excercise from Richard Haberman Mathematical Models, but I don't seems to know what to do to this problem.
mathematical-physics classical-mechanics
mathematical-physics classical-mechanics
asked Jan 22 at 19:33
Niang MooreNiang Moore
326
asked Jan 22 at 19:33
Niang MooreNiang Moore
326
asked Jan 22 at 19:33
Niang MooreNiang Moore
326
asked Jan 22 at 19:33
Niang MooreNiang Moore
326
asked Jan 22 at 19:33
Niang MooreNiang Moore
326
326
$begingroup$
Draw a picture and break position into components.
$endgroup$
– David G. Stork
Jan 22 at 19:37
$begingroup$
What is the definition of simple harmonic motion, in that text?
$endgroup$
– Aretino
Jan 22 at 20:14
add a comment |
$begingroup$
Draw a picture and break position into components.
$endgroup$
– David G. Stork
Jan 22 at 19:37
$begingroup$
What is the definition of simple harmonic motion, in that text?
$endgroup$
– Aretino
Jan 22 at 20:14
$begingroup$
Draw a picture and break position into components.
$endgroup$
– David G. Stork
Jan 22 at 19:37
$begingroup$
Draw a picture and break position into components.
$endgroup$
– David G. Stork
Jan 22 at 19:37
$begingroup$
What is the definition of simple harmonic motion, in that text?
$endgroup$
– Aretino
Jan 22 at 20:14
$begingroup$
What is the definition of simple harmonic motion, in that text?
$endgroup$
– Aretino
Jan 22 at 20:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First we observe that the angle covered, $theta = omega t$
So, we have this.
Intial position of the particle being '$A$' and the current position being '$B$'.
We see that the x-component of the position can be represented as $$mathbf {r cos omega t}$$ where $r$ is the radius of the circle.
It is a $cos$ function and is hence periodic. In fact this equation itself represents that the particle is executing Simple harmonic Motion in the x-component. But you could go ahead and double differentiate it with respect to time and notice that the acceleration is a function of position.
Now, armed with this understanding, I will go ahead and allow the OP to try out the y-component on their own.
Feel free to ask any clarifications!
answered Jan 22 at 19:51
The Jade EmperorThe Jade Emperor
908
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
oldest
votes
$begingroup$
First we observe that the angle covered, $theta = omega t$
So, we have this.
Intial position of the particle being '$A$' and the current position being '$B$'.
We see that the x-component of the position can be represented as $$mathbf {r cos omega t}$$ where $r$ is the radius of the circle.
It is a $cos$ function and is hence periodic. In fact this equation itself represents that the particle is executing Simple harmonic Motion in the x-component. But you could go ahead and double differentiate it with respect to time and notice that the acceleration is a function of position.
Now, armed with this understanding, I will go ahead and allow the OP to try out the y-component on their own.
Feel free to ask any clarifications!
answered Jan 22 at 19:51
The Jade EmperorThe Jade Emperor
908
$endgroup$
add a comment |
$begingroup$
First we observe that the angle covered, $theta = omega t$
So, we have this.
Intial position of the particle being '$A$' and the current position being '$B$'.
We see that the x-component of the position can be represented as $$mathbf {r cos omega t}$$ where $r$ is the radius of the circle.
It is a $cos$ function and is hence periodic. In fact this equation itself represents that the particle is executing Simple harmonic Motion in the x-component. But you could go ahead and double differentiate it with respect to time and notice that the acceleration is a function of position.
Now, armed with this understanding, I will go ahead and allow the OP to try out the y-component on their own.
Feel free to ask any clarifications!
answered Jan 22 at 19:51
The Jade EmperorThe Jade Emperor
908
$endgroup$
add a comment |
$begingroup$
First we observe that the angle covered, $theta = omega t$
So, we have this.
Intial position of the particle being '$A$' and the current position being '$B$'.
We see that the x-component of the position can be represented as $$mathbf {r cos omega t}$$ where $r$ is the radius of the circle.
It is a $cos$ function and is hence periodic. In fact this equation itself represents that the particle is executing Simple harmonic Motion in the x-component. But you could go ahead and double differentiate it with respect to time and notice that the acceleration is a function of position.
Now, armed with this understanding, I will go ahead and allow the OP to try out the y-component on their own.
Feel free to ask any clarifications!
answered Jan 22 at 19:51
The Jade EmperorThe Jade Emperor
908
$endgroup$
First we observe that the angle covered, $theta = omega t$
So, we have this.
Intial position of the particle being '$A$' and the current position being '$B$'.
We see that the x-component of the position can be represented as $$mathbf {r cos omega t}$$ where $r$ is the radius of the circle.
It is a $cos$ function and is hence periodic. In fact this equation itself represents that the particle is executing Simple harmonic Motion in the x-component. But you could go ahead and double differentiate it with respect to time and notice that the acceleration is a function of position.
Now, armed with this understanding, I will go ahead and allow the OP to try out the y-component on their own.
Feel free to ask any clarifications!
answered Jan 22 at 19:51
The Jade EmperorThe Jade Emperor
908
answered Jan 22 at 19:51
The Jade EmperorThe Jade Emperor
908
answered Jan 22 at 19:51
The Jade EmperorThe Jade Emperor
908
answered Jan 22 at 19:51
The Jade EmperorThe Jade Emperor
908
908
add a comment |
add a comment |
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$begingroup$
Draw a picture and break position into components.
$endgroup$
– David G. Stork
Jan 22 at 19:37
$begingroup$
What is the definition of simple harmonic motion, in that text?
$endgroup$
– Aretino
Jan 22 at 20:14