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Why are there two different averages for the kinetic energy in a harmonic oscillator?
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Question:
A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is?
Problem:
For trying to find the answer I used integration to find the average value of the velocity as time dependent function
$$frac12mv^2=frac12mA^2omega^2cos^2omega t,$$
from $0$ to $fracpi2omega$, I am getting $pi^2ma^2nu^2$, which is correct.
However when using velocity as a function of position
$$frac12mv^2=frac12mw^2(A^2-x^2)$$
from $0$ to $A$ I am getting a different answer, $frac13pi^2ma^2nu^2$.
Why is that the case?
harmonic-oscillator
edited 12 mins ago
ACuriousMind♦
68.4k17118293
asked 4 hours ago
Harshit Joshi
3209
add a comment |Â
up vote
4
down vote
favorite
Question:
A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is?
Problem:
For trying to find the answer I used integration to find the average value of the velocity as time dependent function
$$frac12mv^2=frac12mA^2omega^2cos^2omega t,$$
from $0$ to $fracpi2omega$, I am getting $pi^2ma^2nu^2$, which is correct.
However when using velocity as a function of position
$$frac12mv^2=frac12mw^2(A^2-x^2)$$
from $0$ to $A$ I am getting a different answer, $frac13pi^2ma^2nu^2$.
Why is that the case?
harmonic-oscillator
edited 12 mins ago
ACuriousMind♦
68.4k17118293
asked 4 hours ago
Harshit Joshi
3209
The answers above are both wrong, by the way, but you can easily check them and correct them. This doesn't have any impact on my answer below, though.
– LonelyProf
2 hours ago
1
I have reopened this question since it asks a conceptual question which I tried to make clearer by editing the title.
– ACuriousMind♦
10 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Question:
A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is?
Problem:
For trying to find the answer I used integration to find the average value of the velocity as time dependent function
$$frac12mv^2=frac12mA^2omega^2cos^2omega t,$$
from $0$ to $fracpi2omega$, I am getting $pi^2ma^2nu^2$, which is correct.
However when using velocity as a function of position
$$frac12mv^2=frac12mw^2(A^2-x^2)$$
from $0$ to $A$ I am getting a different answer, $frac13pi^2ma^2nu^2$.
Why is that the case?
harmonic-oscillator
edited 12 mins ago
ACuriousMind♦
68.4k17118293
asked 4 hours ago
Harshit Joshi
3209
Question:
A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is?
Problem:
For trying to find the answer I used integration to find the average value of the velocity as time dependent function
$$frac12mv^2=frac12mA^2omega^2cos^2omega t,$$
from $0$ to $fracpi2omega$, I am getting $pi^2ma^2nu^2$, which is correct.
However when using velocity as a function of position
$$frac12mv^2=frac12mw^2(A^2-x^2)$$
from $0$ to $A$ I am getting a different answer, $frac13pi^2ma^2nu^2$.
Why is that the case?
harmonic-oscillator
harmonic-oscillator
edited 12 mins ago
ACuriousMind♦
68.4k17118293
asked 4 hours ago
Harshit Joshi
3209
edited 12 mins ago
ACuriousMind♦
68.4k17118293
asked 4 hours ago
Harshit Joshi
3209
edited 12 mins ago
ACuriousMind♦
68.4k17118293
edited 12 mins ago
ACuriousMind♦
68.4k17118293
edited 12 mins ago
ACuriousMind♦
68.4k17118293
68.4k17118293
asked 4 hours ago
Harshit Joshi
3209
asked 4 hours ago
Harshit Joshi
3209
asked 4 hours ago
Harshit Joshi
3209
3209
The answers above are both wrong, by the way, but you can easily check them and correct them. This doesn't have any impact on my answer below, though.
– LonelyProf
2 hours ago
1
I have reopened this question since it asks a conceptual question which I tried to make clearer by editing the title.
– ACuriousMind♦
10 mins ago
add a comment |Â
The answers above are both wrong, by the way, but you can easily check them and correct them. This doesn't have any impact on my answer below, though.
– LonelyProf
2 hours ago
1
I have reopened this question since it asks a conceptual question which I tried to make clearer by editing the title.
– ACuriousMind♦
10 mins ago
The answers above are both wrong, by the way, but you can easily check them and correct them. This doesn't have any impact on my answer below, though.
– LonelyProf
2 hours ago
The answers above are both wrong, by the way, but you can easily check them and correct them. This doesn't have any impact on my answer below, though.
– LonelyProf
2 hours ago
1
1
I have reopened this question since it asks a conceptual question which I tried to make clearer by editing the title.
– ACuriousMind♦
10 mins ago
I have reopened this question since it asks a conceptual question which I tried to make clearer by editing the title.
– ACuriousMind♦
10 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Although your question is tagged "homework-and-exercises", it is clear that you are not asking us to help you get the right answer, but you are raising a side issue. So I see no problem in offering an answer.
The time-average of $frac12mv^2$ is indeed different from the position-average of the same quantity. There's a different weighting. Actually, any mean value might be defined to include a weighting function, but unless it is specified explicitly, we usually assume an unweighted mean. However, for continuous functions of some variable, the formula for the mean involves integrating with respect to that variable, which implies you have made some choice about the weight.
In your case, you have two obvious candidates. Imagine sampling $N$ discrete values of the kinetic energy at regular intervals of time, and averaging them by adding them up and dividing by $N$. Now imagine that you sample the $N$ values at regularly spaced positions, and again average them in the same way. You would not expect to get the same result: the particle is moving much faster through the position $x=0$, so it spends far less time around that position than around the extremities of the oscillation. The time average will emphasize the lower speeds, compared with the position average.
Explicitly we can write
beginalign*
langle tfrac12mv^2 rangle_x &=
fracint_0^A left(tfrac12mv(x)^2right) dx int_0^A dx
\
textandquad
langle tfrac12mv^2 rangle_t &=
fracint_0^pi/2omega left(tfrac12mv(t)^2right) dt int_0^pi/2omega dt
endalign*
If we change variables in the first equation,
using $dx = v, dt$,
we can see the weight appear explicitly
$$
langle tfrac12mv^2 rangle_x =
fracint_0^pi/2omega v(t), left(tfrac12mv(t)^2right) dt int_0^pi/2omega v(t), dt
$$
The time average is given by a similar expression,
but without the $v$ weighting in the numerator and denominator.
The position average has given more weight to the higher velocities. (Incidentally, if we integrate over a full period,
we need to take a little more care with the sign of the weighting factor)
Either average is "correct". Which one is wanted? Well, the context of the question should really make this clear, but I would probably make the same assumption that you did, and guess that the time-average kinetic energy is likely to be more relevant.
answered 2 hours ago
LonelyProf
1,9432211
add a comment |Â
up vote
-1
down vote
Because $x(t) = A sin omega t$. Then you need to recall your trig identities. Specifically, $sin^2 omega t + cos^2 omega t = 1 $. To get the average KE you need to integrate the KE as a function of $t$ then divide by the change in $t$. OR you need to integrate the KE as a function of $x$ and then divide by the change in $x$.
answered 2 hours ago
puppetsock
1,43037
That's exactly what I did. My question is that why is the position average different from the time average even though both of them refer to the same physical phenomenon?
– Harshit Joshi
2 hours ago
Because you divided by the wrong thing.
– puppetsock
2 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Although your question is tagged "homework-and-exercises", it is clear that you are not asking us to help you get the right answer, but you are raising a side issue. So I see no problem in offering an answer.
The time-average of $frac12mv^2$ is indeed different from the position-average of the same quantity. There's a different weighting. Actually, any mean value might be defined to include a weighting function, but unless it is specified explicitly, we usually assume an unweighted mean. However, for continuous functions of some variable, the formula for the mean involves integrating with respect to that variable, which implies you have made some choice about the weight.
In your case, you have two obvious candidates. Imagine sampling $N$ discrete values of the kinetic energy at regular intervals of time, and averaging them by adding them up and dividing by $N$. Now imagine that you sample the $N$ values at regularly spaced positions, and again average them in the same way. You would not expect to get the same result: the particle is moving much faster through the position $x=0$, so it spends far less time around that position than around the extremities of the oscillation. The time average will emphasize the lower speeds, compared with the position average.
Explicitly we can write
beginalign*
langle tfrac12mv^2 rangle_x &=
fracint_0^A left(tfrac12mv(x)^2right) dx int_0^A dx
\
textandquad
langle tfrac12mv^2 rangle_t &=
fracint_0^pi/2omega left(tfrac12mv(t)^2right) dt int_0^pi/2omega dt
endalign*
If we change variables in the first equation,
using $dx = v, dt$,
we can see the weight appear explicitly
$$
langle tfrac12mv^2 rangle_x =
fracint_0^pi/2omega v(t), left(tfrac12mv(t)^2right) dt int_0^pi/2omega v(t), dt
$$
The time average is given by a similar expression,
but without the $v$ weighting in the numerator and denominator.
The position average has given more weight to the higher velocities. (Incidentally, if we integrate over a full period,
we need to take a little more care with the sign of the weighting factor)
Either average is "correct". Which one is wanted? Well, the context of the question should really make this clear, but I would probably make the same assumption that you did, and guess that the time-average kinetic energy is likely to be more relevant.
answered 2 hours ago
LonelyProf
1,9432211
add a comment |Â
up vote
4
down vote
accepted
Although your question is tagged "homework-and-exercises", it is clear that you are not asking us to help you get the right answer, but you are raising a side issue. So I see no problem in offering an answer.
The time-average of $frac12mv^2$ is indeed different from the position-average of the same quantity. There's a different weighting. Actually, any mean value might be defined to include a weighting function, but unless it is specified explicitly, we usually assume an unweighted mean. However, for continuous functions of some variable, the formula for the mean involves integrating with respect to that variable, which implies you have made some choice about the weight.
In your case, you have two obvious candidates. Imagine sampling $N$ discrete values of the kinetic energy at regular intervals of time, and averaging them by adding them up and dividing by $N$. Now imagine that you sample the $N$ values at regularly spaced positions, and again average them in the same way. You would not expect to get the same result: the particle is moving much faster through the position $x=0$, so it spends far less time around that position than around the extremities of the oscillation. The time average will emphasize the lower speeds, compared with the position average.
Explicitly we can write
beginalign*
langle tfrac12mv^2 rangle_x &=
fracint_0^A left(tfrac12mv(x)^2right) dx int_0^A dx
\
textandquad
langle tfrac12mv^2 rangle_t &=
fracint_0^pi/2omega left(tfrac12mv(t)^2right) dt int_0^pi/2omega dt
endalign*
If we change variables in the first equation,
using $dx = v, dt$,
we can see the weight appear explicitly
$$
langle tfrac12mv^2 rangle_x =
fracint_0^pi/2omega v(t), left(tfrac12mv(t)^2right) dt int_0^pi/2omega v(t), dt
$$
The time average is given by a similar expression,
but without the $v$ weighting in the numerator and denominator.
The position average has given more weight to the higher velocities. (Incidentally, if we integrate over a full period,
we need to take a little more care with the sign of the weighting factor)
Either average is "correct". Which one is wanted? Well, the context of the question should really make this clear, but I would probably make the same assumption that you did, and guess that the time-average kinetic energy is likely to be more relevant.
answered 2 hours ago
LonelyProf
1,9432211
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Although your question is tagged "homework-and-exercises", it is clear that you are not asking us to help you get the right answer, but you are raising a side issue. So I see no problem in offering an answer.
The time-average of $frac12mv^2$ is indeed different from the position-average of the same quantity. There's a different weighting. Actually, any mean value might be defined to include a weighting function, but unless it is specified explicitly, we usually assume an unweighted mean. However, for continuous functions of some variable, the formula for the mean involves integrating with respect to that variable, which implies you have made some choice about the weight.
In your case, you have two obvious candidates. Imagine sampling $N$ discrete values of the kinetic energy at regular intervals of time, and averaging them by adding them up and dividing by $N$. Now imagine that you sample the $N$ values at regularly spaced positions, and again average them in the same way. You would not expect to get the same result: the particle is moving much faster through the position $x=0$, so it spends far less time around that position than around the extremities of the oscillation. The time average will emphasize the lower speeds, compared with the position average.
Explicitly we can write
beginalign*
langle tfrac12mv^2 rangle_x &=
fracint_0^A left(tfrac12mv(x)^2right) dx int_0^A dx
\
textandquad
langle tfrac12mv^2 rangle_t &=
fracint_0^pi/2omega left(tfrac12mv(t)^2right) dt int_0^pi/2omega dt
endalign*
If we change variables in the first equation,
using $dx = v, dt$,
we can see the weight appear explicitly
$$
langle tfrac12mv^2 rangle_x =
fracint_0^pi/2omega v(t), left(tfrac12mv(t)^2right) dt int_0^pi/2omega v(t), dt
$$
The time average is given by a similar expression,
but without the $v$ weighting in the numerator and denominator.
The position average has given more weight to the higher velocities. (Incidentally, if we integrate over a full period,
we need to take a little more care with the sign of the weighting factor)
Either average is "correct". Which one is wanted? Well, the context of the question should really make this clear, but I would probably make the same assumption that you did, and guess that the time-average kinetic energy is likely to be more relevant.
answered 2 hours ago
LonelyProf
1,9432211
Although your question is tagged "homework-and-exercises", it is clear that you are not asking us to help you get the right answer, but you are raising a side issue. So I see no problem in offering an answer.
The time-average of $frac12mv^2$ is indeed different from the position-average of the same quantity. There's a different weighting. Actually, any mean value might be defined to include a weighting function, but unless it is specified explicitly, we usually assume an unweighted mean. However, for continuous functions of some variable, the formula for the mean involves integrating with respect to that variable, which implies you have made some choice about the weight.
In your case, you have two obvious candidates. Imagine sampling $N$ discrete values of the kinetic energy at regular intervals of time, and averaging them by adding them up and dividing by $N$. Now imagine that you sample the $N$ values at regularly spaced positions, and again average them in the same way. You would not expect to get the same result: the particle is moving much faster through the position $x=0$, so it spends far less time around that position than around the extremities of the oscillation. The time average will emphasize the lower speeds, compared with the position average.
Explicitly we can write
beginalign*
langle tfrac12mv^2 rangle_x &=
fracint_0^A left(tfrac12mv(x)^2right) dx int_0^A dx
\
textandquad
langle tfrac12mv^2 rangle_t &=
fracint_0^pi/2omega left(tfrac12mv(t)^2right) dt int_0^pi/2omega dt
endalign*
If we change variables in the first equation,
using $dx = v, dt$,
we can see the weight appear explicitly
$$
langle tfrac12mv^2 rangle_x =
fracint_0^pi/2omega v(t), left(tfrac12mv(t)^2right) dt int_0^pi/2omega v(t), dt
$$
The time average is given by a similar expression,
but without the $v$ weighting in the numerator and denominator.
The position average has given more weight to the higher velocities. (Incidentally, if we integrate over a full period,
we need to take a little more care with the sign of the weighting factor)
Either average is "correct". Which one is wanted? Well, the context of the question should really make this clear, but I would probably make the same assumption that you did, and guess that the time-average kinetic energy is likely to be more relevant.
answered 2 hours ago
LonelyProf
1,9432211
answered 2 hours ago
LonelyProf
1,9432211
answered 2 hours ago
LonelyProf
1,9432211
answered 2 hours ago
LonelyProf
1,9432211
1,9432211
add a comment |Â
add a comment |Â
up vote
-1
down vote
Because $x(t) = A sin omega t$. Then you need to recall your trig identities. Specifically, $sin^2 omega t + cos^2 omega t = 1 $. To get the average KE you need to integrate the KE as a function of $t$ then divide by the change in $t$. OR you need to integrate the KE as a function of $x$ and then divide by the change in $x$.
answered 2 hours ago
puppetsock
1,43037
That's exactly what I did. My question is that why is the position average different from the time average even though both of them refer to the same physical phenomenon?
– Harshit Joshi
2 hours ago
Because you divided by the wrong thing.
– puppetsock
2 hours ago
add a comment |Â
up vote
-1
down vote
Because $x(t) = A sin omega t$. Then you need to recall your trig identities. Specifically, $sin^2 omega t + cos^2 omega t = 1 $. To get the average KE you need to integrate the KE as a function of $t$ then divide by the change in $t$. OR you need to integrate the KE as a function of $x$ and then divide by the change in $x$.
answered 2 hours ago
puppetsock
1,43037
That's exactly what I did. My question is that why is the position average different from the time average even though both of them refer to the same physical phenomenon?
– Harshit Joshi
2 hours ago
Because you divided by the wrong thing.
– puppetsock
2 hours ago
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Because $x(t) = A sin omega t$. Then you need to recall your trig identities. Specifically, $sin^2 omega t + cos^2 omega t = 1 $. To get the average KE you need to integrate the KE as a function of $t$ then divide by the change in $t$. OR you need to integrate the KE as a function of $x$ and then divide by the change in $x$.
answered 2 hours ago
puppetsock
1,43037
Because $x(t) = A sin omega t$. Then you need to recall your trig identities. Specifically, $sin^2 omega t + cos^2 omega t = 1 $. To get the average KE you need to integrate the KE as a function of $t$ then divide by the change in $t$. OR you need to integrate the KE as a function of $x$ and then divide by the change in $x$.
answered 2 hours ago
puppetsock
1,43037
answered 2 hours ago
puppetsock
1,43037
answered 2 hours ago
puppetsock
1,43037
answered 2 hours ago
puppetsock
1,43037
1,43037
That's exactly what I did. My question is that why is the position average different from the time average even though both of them refer to the same physical phenomenon?
– Harshit Joshi
2 hours ago
Because you divided by the wrong thing.
– puppetsock
2 hours ago
add a comment |Â
That's exactly what I did. My question is that why is the position average different from the time average even though both of them refer to the same physical phenomenon?
– Harshit Joshi
2 hours ago
Because you divided by the wrong thing.
– puppetsock
2 hours ago
That's exactly what I did. My question is that why is the position average different from the time average even though both of them refer to the same physical phenomenon?
– Harshit Joshi
2 hours ago
That's exactly what I did. My question is that why is the position average different from the time average even though both of them refer to the same physical phenomenon?
– Harshit Joshi
2 hours ago
Because you divided by the wrong thing.
– puppetsock
2 hours ago
Because you divided by the wrong thing.
– puppetsock
2 hours ago
add a comment |Â
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The answers above are both wrong, by the way, but you can easily check them and correct them. This doesn't have any impact on my answer below, though.
– LonelyProf
2 hours ago
1
I have reopened this question since it asks a conceptual question which I tried to make clearer by editing the title.
– ACuriousMind♦
10 mins ago