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Solving forced undamped vibration using Laplace transforms
$begingroup$
I'm heaving trouble solving the following undamped forced vibration problem using Laplace transforms:
$$ddot{q}(t) + omega_n^2 q(t) = cos(omega t).$$
I will show what I have done so far, and I'd appreciate any insights.
Taking the Laplace transform of both sides and applying the derivative identities yields,
$$mathcal{L}{ddot{q}(t)} + mathcal{L}{omega_n^2 q(t)} = mathcal{L}{cos(omega t)},$$
$$s^2 mathcal{L}{{q}(t)} - s q(0) - dot{q}(0) + omega_n^2 mathcal{L}{q(t)} = frac{s}{s^2 + omega^2},$$
I asume the initial conditions are zero, then group and solve for $mathcal{L}{{q}(t)}$,
$$mathcal{L}{{q}(t)} = frac{1}{s^2 + omega_n^2} frac{s}{s^2 + omega^2} .$$
Multiplying and dividing by $omega_n$ allows recognizing the convolution product,
$$ mathcal{L}{{q}(t)} = frac{1}{omega_n} mathcal{L}{sin(omega_n t)} mathcal{L}{cos(omega t)}.$$
And the time-domain solution would be
$$
q(t) = frac{1}{omega_n} int_{0}^{t} cos(omega (t - tau))sin(omega_n tau) ,d tau.
$$
Now here is my problem: I don't know if made a mistake up to this point, but assuming I haven't when I try to evaluate the convolution integral the results are not what I expected, namely I expected,
$$
lim_{omega rightarrow omega_n} lim_{t rightarrow infty} q(t) rightarrow infty,
$$
but I don't get that from the convolution integral.
Any thought?
ordinary-differential-equations laplace-transform convolution
asked Sep 27 '15 at 17:15
Indrid ColdIndrid Cold
9717
$endgroup$
|
show 7 more comments
$begingroup$
I'm heaving trouble solving the following undamped forced vibration problem using Laplace transforms:
$$ddot{q}(t) + omega_n^2 q(t) = cos(omega t).$$
I will show what I have done so far, and I'd appreciate any insights.
Taking the Laplace transform of both sides and applying the derivative identities yields,
$$mathcal{L}{ddot{q}(t)} + mathcal{L}{omega_n^2 q(t)} = mathcal{L}{cos(omega t)},$$
$$s^2 mathcal{L}{{q}(t)} - s q(0) - dot{q}(0) + omega_n^2 mathcal{L}{q(t)} = frac{s}{s^2 + omega^2},$$
I asume the initial conditions are zero, then group and solve for $mathcal{L}{{q}(t)}$,
$$mathcal{L}{{q}(t)} = frac{1}{s^2 + omega_n^2} frac{s}{s^2 + omega^2} .$$
Multiplying and dividing by $omega_n$ allows recognizing the convolution product,
$$ mathcal{L}{{q}(t)} = frac{1}{omega_n} mathcal{L}{sin(omega_n t)} mathcal{L}{cos(omega t)}.$$
And the time-domain solution would be
$$
q(t) = frac{1}{omega_n} int_{0}^{t} cos(omega (t - tau))sin(omega_n tau) ,d tau.
$$
Now here is my problem: I don't know if made a mistake up to this point, but assuming I haven't when I try to evaluate the convolution integral the results are not what I expected, namely I expected,
$$
lim_{omega rightarrow omega_n} lim_{t rightarrow infty} q(t) rightarrow infty,
$$
but I don't get that from the convolution integral.
Any thought?
ordinary-differential-equations laplace-transform convolution
asked Sep 27 '15 at 17:15
Indrid ColdIndrid Cold
9717
$endgroup$
$begingroup$
You should anticipate that the amplitude of the oscillation grows over time, not that the limit is $+infty$.
$endgroup$
– Ian
Sep 27 '15 at 18:24
1
$begingroup$
In fact being explicit your solution is apparently $frac{cos(omega t)-cos(omega_n t)}{omega_n^2-omega^2}$, or something like this. For $t gg 1/|omega_n-omega|$, the numerator will be (sometimes) on the order of $1$ while the denominator will remain small.
$endgroup$
– Ian
Sep 27 '15 at 18:31
$begingroup$
You can qualitatively see what is going on when you look at $omega=omega_n$, then the solution is $t$ times a sinusoid (maybe plus some periodic terms). So it does not blow up monotonically but the amplitude still grows. For $omega$ close to $omega_n$ the amplitude grows but not without bound.
$endgroup$
– Ian
Sep 27 '15 at 18:32
$begingroup$
Ian, you actually answered the question. I evaluated the convolution integral incorrectly. And indeed the solution tends to infinity as $omega rightarrow omega_n$.
$endgroup$
– Indrid Cold
Sep 27 '15 at 18:49
$begingroup$
...No, it doesn't. Again, look at the truly resonant case: $y''+y=cos(y),y(0)=0,y'(0)=0$ has solution $y=frac{1}{2} t sin(t)$. On a fixed finite time interval, the solution where we replace $cos(y)$ with $cos(ay)$ for $a approx 1$ will be similar to this.
$endgroup$
– Ian
Sep 27 '15 at 18:51
|
show 7 more comments
$begingroup$
I'm heaving trouble solving the following undamped forced vibration problem using Laplace transforms:
$$ddot{q}(t) + omega_n^2 q(t) = cos(omega t).$$
I will show what I have done so far, and I'd appreciate any insights.
Taking the Laplace transform of both sides and applying the derivative identities yields,
$$mathcal{L}{ddot{q}(t)} + mathcal{L}{omega_n^2 q(t)} = mathcal{L}{cos(omega t)},$$
$$s^2 mathcal{L}{{q}(t)} - s q(0) - dot{q}(0) + omega_n^2 mathcal{L}{q(t)} = frac{s}{s^2 + omega^2},$$
I asume the initial conditions are zero, then group and solve for $mathcal{L}{{q}(t)}$,
$$mathcal{L}{{q}(t)} = frac{1}{s^2 + omega_n^2} frac{s}{s^2 + omega^2} .$$
Multiplying and dividing by $omega_n$ allows recognizing the convolution product,
$$ mathcal{L}{{q}(t)} = frac{1}{omega_n} mathcal{L}{sin(omega_n t)} mathcal{L}{cos(omega t)}.$$
And the time-domain solution would be
$$
q(t) = frac{1}{omega_n} int_{0}^{t} cos(omega (t - tau))sin(omega_n tau) ,d tau.
$$
Now here is my problem: I don't know if made a mistake up to this point, but assuming I haven't when I try to evaluate the convolution integral the results are not what I expected, namely I expected,
$$
lim_{omega rightarrow omega_n} lim_{t rightarrow infty} q(t) rightarrow infty,
$$
but I don't get that from the convolution integral.
Any thought?
ordinary-differential-equations laplace-transform convolution
asked Sep 27 '15 at 17:15
Indrid ColdIndrid Cold
9717
$endgroup$
I'm heaving trouble solving the following undamped forced vibration problem using Laplace transforms:
$$ddot{q}(t) + omega_n^2 q(t) = cos(omega t).$$
I will show what I have done so far, and I'd appreciate any insights.
Taking the Laplace transform of both sides and applying the derivative identities yields,
$$mathcal{L}{ddot{q}(t)} + mathcal{L}{omega_n^2 q(t)} = mathcal{L}{cos(omega t)},$$
$$s^2 mathcal{L}{{q}(t)} - s q(0) - dot{q}(0) + omega_n^2 mathcal{L}{q(t)} = frac{s}{s^2 + omega^2},$$
I asume the initial conditions are zero, then group and solve for $mathcal{L}{{q}(t)}$,
$$mathcal{L}{{q}(t)} = frac{1}{s^2 + omega_n^2} frac{s}{s^2 + omega^2} .$$
Multiplying and dividing by $omega_n$ allows recognizing the convolution product,
$$ mathcal{L}{{q}(t)} = frac{1}{omega_n} mathcal{L}{sin(omega_n t)} mathcal{L}{cos(omega t)}.$$
And the time-domain solution would be
$$
q(t) = frac{1}{omega_n} int_{0}^{t} cos(omega (t - tau))sin(omega_n tau) ,d tau.
$$
Now here is my problem: I don't know if made a mistake up to this point, but assuming I haven't when I try to evaluate the convolution integral the results are not what I expected, namely I expected,
$$
lim_{omega rightarrow omega_n} lim_{t rightarrow infty} q(t) rightarrow infty,
$$
but I don't get that from the convolution integral.
Any thought?
ordinary-differential-equations laplace-transform convolution
ordinary-differential-equations laplace-transform convolution
asked Sep 27 '15 at 17:15
Indrid ColdIndrid Cold
9717
asked Sep 27 '15 at 17:15
Indrid ColdIndrid Cold
9717
edited Sep 27 '15 at 20:00
Indrid Cold
asked Sep 27 '15 at 17:15
Indrid ColdIndrid Cold
9717
asked Sep 27 '15 at 17:15
Indrid ColdIndrid Cold
9717
asked Sep 27 '15 at 17:15
Indrid ColdIndrid Cold
9717
9717
$begingroup$
You should anticipate that the amplitude of the oscillation grows over time, not that the limit is $+infty$.
$endgroup$
– Ian
Sep 27 '15 at 18:24
1
$begingroup$
In fact being explicit your solution is apparently $frac{cos(omega t)-cos(omega_n t)}{omega_n^2-omega^2}$, or something like this. For $t gg 1/|omega_n-omega|$, the numerator will be (sometimes) on the order of $1$ while the denominator will remain small.
$endgroup$
– Ian
Sep 27 '15 at 18:31
$begingroup$
You can qualitatively see what is going on when you look at $omega=omega_n$, then the solution is $t$ times a sinusoid (maybe plus some periodic terms). So it does not blow up monotonically but the amplitude still grows. For $omega$ close to $omega_n$ the amplitude grows but not without bound.
$endgroup$
– Ian
Sep 27 '15 at 18:32
$begingroup$
Ian, you actually answered the question. I evaluated the convolution integral incorrectly. And indeed the solution tends to infinity as $omega rightarrow omega_n$.
$endgroup$
– Indrid Cold
Sep 27 '15 at 18:49
$begingroup$
...No, it doesn't. Again, look at the truly resonant case: $y''+y=cos(y),y(0)=0,y'(0)=0$ has solution $y=frac{1}{2} t sin(t)$. On a fixed finite time interval, the solution where we replace $cos(y)$ with $cos(ay)$ for $a approx 1$ will be similar to this.
$endgroup$
– Ian
Sep 27 '15 at 18:51
|
show 7 more comments
$begingroup$
You should anticipate that the amplitude of the oscillation grows over time, not that the limit is $+infty$.
$endgroup$
– Ian
Sep 27 '15 at 18:24
1
$begingroup$
In fact being explicit your solution is apparently $frac{cos(omega t)-cos(omega_n t)}{omega_n^2-omega^2}$, or something like this. For $t gg 1/|omega_n-omega|$, the numerator will be (sometimes) on the order of $1$ while the denominator will remain small.
$endgroup$
– Ian
Sep 27 '15 at 18:31
$begingroup$
You can qualitatively see what is going on when you look at $omega=omega_n$, then the solution is $t$ times a sinusoid (maybe plus some periodic terms). So it does not blow up monotonically but the amplitude still grows. For $omega$ close to $omega_n$ the amplitude grows but not without bound.
$endgroup$
– Ian
Sep 27 '15 at 18:32
$begingroup$
Ian, you actually answered the question. I evaluated the convolution integral incorrectly. And indeed the solution tends to infinity as $omega rightarrow omega_n$.
$endgroup$
– Indrid Cold
Sep 27 '15 at 18:49
$begingroup$
...No, it doesn't. Again, look at the truly resonant case: $y''+y=cos(y),y(0)=0,y'(0)=0$ has solution $y=frac{1}{2} t sin(t)$. On a fixed finite time interval, the solution where we replace $cos(y)$ with $cos(ay)$ for $a approx 1$ will be similar to this.
$endgroup$
– Ian
Sep 27 '15 at 18:51
$begingroup$
You should anticipate that the amplitude of the oscillation grows over time, not that the limit is $+infty$.
$endgroup$
– Ian
Sep 27 '15 at 18:24
$begingroup$
You should anticipate that the amplitude of the oscillation grows over time, not that the limit is $+infty$.
$endgroup$
– Ian
Sep 27 '15 at 18:24
1
1
$begingroup$
In fact being explicit your solution is apparently $frac{cos(omega t)-cos(omega_n t)}{omega_n^2-omega^2}$, or something like this. For $t gg 1/|omega_n-omega|$, the numerator will be (sometimes) on the order of $1$ while the denominator will remain small.
$endgroup$
– Ian
Sep 27 '15 at 18:31
$begingroup$
In fact being explicit your solution is apparently $frac{cos(omega t)-cos(omega_n t)}{omega_n^2-omega^2}$, or something like this. For $t gg 1/|omega_n-omega|$, the numerator will be (sometimes) on the order of $1$ while the denominator will remain small.
$endgroup$
– Ian
Sep 27 '15 at 18:31
$begingroup$
You can qualitatively see what is going on when you look at $omega=omega_n$, then the solution is $t$ times a sinusoid (maybe plus some periodic terms). So it does not blow up monotonically but the amplitude still grows. For $omega$ close to $omega_n$ the amplitude grows but not without bound.
$endgroup$
– Ian
Sep 27 '15 at 18:32
$begingroup$
You can qualitatively see what is going on when you look at $omega=omega_n$, then the solution is $t$ times a sinusoid (maybe plus some periodic terms). So it does not blow up monotonically but the amplitude still grows. For $omega$ close to $omega_n$ the amplitude grows but not without bound.
$endgroup$
– Ian
Sep 27 '15 at 18:32
$begingroup$
Ian, you actually answered the question. I evaluated the convolution integral incorrectly. And indeed the solution tends to infinity as $omega rightarrow omega_n$.
$endgroup$
– Indrid Cold
Sep 27 '15 at 18:49
$begingroup$
Ian, you actually answered the question. I evaluated the convolution integral incorrectly. And indeed the solution tends to infinity as $omega rightarrow omega_n$.
$endgroup$
– Indrid Cold
Sep 27 '15 at 18:49
$begingroup$
...No, it doesn't. Again, look at the truly resonant case: $y''+y=cos(y),y(0)=0,y'(0)=0$ has solution $y=frac{1}{2} t sin(t)$. On a fixed finite time interval, the solution where we replace $cos(y)$ with $cos(ay)$ for $a approx 1$ will be similar to this.
$endgroup$
– Ian
Sep 27 '15 at 18:51
$begingroup$
...No, it doesn't. Again, look at the truly resonant case: $y''+y=cos(y),y(0)=0,y'(0)=0$ has solution $y=frac{1}{2} t sin(t)$. On a fixed finite time interval, the solution where we replace $cos(y)$ with $cos(ay)$ for $a approx 1$ will be similar to this.
$endgroup$
– Ian
Sep 27 '15 at 18:51
|
show 7 more comments
1 Answer
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$begingroup$
I had evaluated the convolution integral incorrectly. The answer is
$$
q(t) = frac{1}{omega_n} int_{0}^{t} cos(omega(t-tau)) sin(omega_n tau) ,d tau = frac{cos(omega t) - cos(omega_n t)}{omega_n^2 - omega^2}
$$
The limit as $omega rightarrow omega_n$ can be obtained L'Hopital's rule,
$$lim_{omega rightarrow omega_n} q(t) = frac{t sin(omega_n t)}{2 omega_n}$$
Clearly the magnitude of the response approaches infinity as $t rightarrow infty$.
Thanks to Ian for his comments.
answered Sep 27 '15 at 20:04
Indrid ColdIndrid Cold
9717
$endgroup$
add a comment |
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$begingroup$
I had evaluated the convolution integral incorrectly. The answer is
$$
q(t) = frac{1}{omega_n} int_{0}^{t} cos(omega(t-tau)) sin(omega_n tau) ,d tau = frac{cos(omega t) - cos(omega_n t)}{omega_n^2 - omega^2}
$$
The limit as $omega rightarrow omega_n$ can be obtained L'Hopital's rule,
$$lim_{omega rightarrow omega_n} q(t) = frac{t sin(omega_n t)}{2 omega_n}$$
Clearly the magnitude of the response approaches infinity as $t rightarrow infty$.
Thanks to Ian for his comments.
answered Sep 27 '15 at 20:04
Indrid ColdIndrid Cold
9717
$endgroup$
add a comment |
$begingroup$
I had evaluated the convolution integral incorrectly. The answer is
$$
q(t) = frac{1}{omega_n} int_{0}^{t} cos(omega(t-tau)) sin(omega_n tau) ,d tau = frac{cos(omega t) - cos(omega_n t)}{omega_n^2 - omega^2}
$$
The limit as $omega rightarrow omega_n$ can be obtained L'Hopital's rule,
$$lim_{omega rightarrow omega_n} q(t) = frac{t sin(omega_n t)}{2 omega_n}$$
Clearly the magnitude of the response approaches infinity as $t rightarrow infty$.
Thanks to Ian for his comments.
answered Sep 27 '15 at 20:04
Indrid ColdIndrid Cold
9717
$endgroup$
add a comment |
$begingroup$
I had evaluated the convolution integral incorrectly. The answer is
$$
q(t) = frac{1}{omega_n} int_{0}^{t} cos(omega(t-tau)) sin(omega_n tau) ,d tau = frac{cos(omega t) - cos(omega_n t)}{omega_n^2 - omega^2}
$$
The limit as $omega rightarrow omega_n$ can be obtained L'Hopital's rule,
$$lim_{omega rightarrow omega_n} q(t) = frac{t sin(omega_n t)}{2 omega_n}$$
Clearly the magnitude of the response approaches infinity as $t rightarrow infty$.
Thanks to Ian for his comments.
answered Sep 27 '15 at 20:04
Indrid ColdIndrid Cold
9717
$endgroup$
I had evaluated the convolution integral incorrectly. The answer is
$$
q(t) = frac{1}{omega_n} int_{0}^{t} cos(omega(t-tau)) sin(omega_n tau) ,d tau = frac{cos(omega t) - cos(omega_n t)}{omega_n^2 - omega^2}
$$
The limit as $omega rightarrow omega_n$ can be obtained L'Hopital's rule,
$$lim_{omega rightarrow omega_n} q(t) = frac{t sin(omega_n t)}{2 omega_n}$$
Clearly the magnitude of the response approaches infinity as $t rightarrow infty$.
Thanks to Ian for his comments.
answered Sep 27 '15 at 20:04
Indrid ColdIndrid Cold
9717
answered Sep 27 '15 at 20:04
Indrid ColdIndrid Cold
9717
answered Sep 27 '15 at 20:04
Indrid ColdIndrid Cold
9717
answered Sep 27 '15 at 20:04
Indrid ColdIndrid Cold
9717
9717
add a comment |
add a comment |
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$begingroup$
You should anticipate that the amplitude of the oscillation grows over time, not that the limit is $+infty$.
$endgroup$
– Ian
Sep 27 '15 at 18:24
1
$begingroup$
In fact being explicit your solution is apparently $frac{cos(omega t)-cos(omega_n t)}{omega_n^2-omega^2}$, or something like this. For $t gg 1/|omega_n-omega|$, the numerator will be (sometimes) on the order of $1$ while the denominator will remain small.
$endgroup$
– Ian
Sep 27 '15 at 18:31
$begingroup$
You can qualitatively see what is going on when you look at $omega=omega_n$, then the solution is $t$ times a sinusoid (maybe plus some periodic terms). So it does not blow up monotonically but the amplitude still grows. For $omega$ close to $omega_n$ the amplitude grows but not without bound.
$endgroup$
– Ian
Sep 27 '15 at 18:32
$begingroup$
Ian, you actually answered the question. I evaluated the convolution integral incorrectly. And indeed the solution tends to infinity as $omega rightarrow omega_n$.
$endgroup$
– Indrid Cold
Sep 27 '15 at 18:49
$begingroup$
...No, it doesn't. Again, look at the truly resonant case: $y''+y=cos(y),y(0)=0,y'(0)=0$ has solution $y=frac{1}{2} t sin(t)$. On a fixed finite time interval, the solution where we replace $cos(y)$ with $cos(ay)$ for $a approx 1$ will be similar to this.
$endgroup$
– Ian
Sep 27 '15 at 18:51