This is equal to the 2nd derivative of the impulse function. Can we realize this physically? I mean, is there any use for this in the real world?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestWell, inverse Laplace transforms aside, if the question is about the physical meaning of the 2nd derivative of the impulse function, which can be approximated with a Gaussian function, then we can propose this chart, where \(J\) is the impulse function

\(J\) = integral of force over time, or change in momentum

\(J'\) = measure of rise (or fall) of force over time (this would be "jerk" in mechanics)

\(J''\) = measure of acceleration (or de-acceleration) of force over time (this should not be confused with ordinary acceleration--no forces are associated with this term)

The more "pointy" the Gaussian function becomes, the more it becomes like the true impulse or Dirac Delta function, so we could use this as a kind of an index for this. That is, if

\(J=\dfrac { 1 }{ a\sqrt { \pi } } { e }^{ -\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } }\)

then

\(J''(0)=-\dfrac { 2 }{ { a }^{ 2 }\sqrt { \pi } } \)

So we just have a knob, \(a\), in which we can vary from a continuous application of force to the special case where the application of force is discrete and instantaneous. – Michael Mendrin · 2 years, 4 months ago

Log in to reply