- #1
thegreenlaser
- 525
- 16
I'm reading through some material on vibrating strings. In one book, the author repeatedly uses energy analysis to determine information about a system. What I'm not understanding is how he gets his energy equations. For example, with the equation:
[tex]\frac{d^2 u}{dt^2}=-\omega_0^2 u[/tex]
He finds the energy by multiplying the whole equation by the velocity,
[tex]\frac{du}{dt}\frac{d^2 u}{dt^2} + \omega_0^2 \frac{du}{dt}u = 0[/tex]
and then pulling out the derivative, or kind of semi-integrating:
[tex]\frac{d}{dt}\left( \frac{1}{2} {\left(\frac{du}{dt}\right)}^2 + \frac{\omega_0^2}{2} u^2 \right) = 0[/tex]
So now he says that this represents the derivative of the energy, scaled by a constant with units of mass. i.e. the above equation means that the time derivative of the energy in the system is 0.
I can see the kinetic and potential energy terms in there, but I'm really not understanding this 'multiplying by du/dt' technique. Can anyone explain the justification for this? He uses it constantly throughout the book, and when it gets to partial derivatives he starts using an inner product defined by:
[tex]\left< f,g \right>_D = \int_D f \cdot g \ dt[/tex]
So to find the derivative of the energy of the system, he takes the inner product of the whole equation with the time derivative of u, and then pulls out the differential operator similar to above to obtain something like (partial / partial t)(expression) = 0.
I haven't done any physics to do with energy since high school, so I might be missing something obvious, but can anyone explain why multiplying by velocity (or more generally taking the inner product with velocity) will yield the derivative of energy scaled by mass?
[tex]\frac{d^2 u}{dt^2}=-\omega_0^2 u[/tex]
He finds the energy by multiplying the whole equation by the velocity,
[tex]\frac{du}{dt}\frac{d^2 u}{dt^2} + \omega_0^2 \frac{du}{dt}u = 0[/tex]
and then pulling out the derivative, or kind of semi-integrating:
[tex]\frac{d}{dt}\left( \frac{1}{2} {\left(\frac{du}{dt}\right)}^2 + \frac{\omega_0^2}{2} u^2 \right) = 0[/tex]
So now he says that this represents the derivative of the energy, scaled by a constant with units of mass. i.e. the above equation means that the time derivative of the energy in the system is 0.
I can see the kinetic and potential energy terms in there, but I'm really not understanding this 'multiplying by du/dt' technique. Can anyone explain the justification for this? He uses it constantly throughout the book, and when it gets to partial derivatives he starts using an inner product defined by:
[tex]\left< f,g \right>_D = \int_D f \cdot g \ dt[/tex]
So to find the derivative of the energy of the system, he takes the inner product of the whole equation with the time derivative of u, and then pulls out the differential operator similar to above to obtain something like (partial / partial t)(expression) = 0.
I haven't done any physics to do with energy since high school, so I might be missing something obvious, but can anyone explain why multiplying by velocity (or more generally taking the inner product with velocity) will yield the derivative of energy scaled by mass?