Solve Time Rate of Change of Mechanical Energy for Damped, Undriven Oscillator

AI Thread Summary
The time rate of change of mechanical energy for a damped, undriven oscillator is expressed as dE/dt = -bv^2, indicating that energy is always decreasing. To derive this, the mechanical energy equation E = 1/2mv^2 + 1/2kx^2 is differentiated with respect to time. Utilizing the equation -kx - b(dx/dt) = m(d^2x/dt^2), the differentiation process incorporates both velocity and acceleration. After simplification, it confirms that the energy loss is due to damping forces, which is consistent with the behavior of damped oscillators. This analysis highlights the inherent energy dissipation in such systems.
Soilwork
Messages
101
Reaction score
0
Hey can someone guide me in the right direction here.

Q. Show that the time rate of change of mechanical energy for a damped , ubdriven oscillator is given by dE/dt=-bv^2 and hence is always negative.
Proceed as follows: Differentiate the expression for the mechanical energy of an oscillator, E=1/2mv^2 + 1/2kx^2. And use this equation:
-kx-b(dx/dt)=m(d^2x/dt^2)
I know it tells you how to approach the question and all, but I guess the main problem is that my differentiation isn't that great.
Anyway for the Energy equation I was thinking that you could change the v to change in x over change in time. delta(x)/delta(t)
not sure though.
Any help would be appreciated.
 
Physics news on Phys.org
nevermind :)
It was easy, it's just that I made a small error at the beginning that through me off a bit.
 


To solve for the time rate of change of mechanical energy for a damped, undriven oscillator, we can use the expression for mechanical energy, E=1/2mv^2+1/2kx^2, and differentiate it with respect to time. This will give us:

dE/dt = d/dt(1/2mv^2) + d/dt(1/2kx^2)

Using the chain rule, we can rewrite this as:

dE/dt = (1/2m)(d/dt(v^2)) + (1/2k)(d/dt(x^2))

Now, we can use the equation given in the problem, -kx-b(dx/dt)=m(d^2x/dt^2), to replace the second term in the above equation. This will give us:

dE/dt = (1/2m)(d/dt(v^2)) + (1/2k)(-kx-b(dx/dt))

Simplifying this, we get:

dE/dt = (1/2m)(2v)(dv/dt) + (-kx-b(dx/dt))

Using the fact that dv/dt = a (acceleration), and dx/dt = v (velocity), we can rewrite this as:

dE/dt = mv(a) + (-kx-bv)

Simplifying further, we get:

dE/dt = mav + (-kx-bv)

Now, we can use the equation for acceleration, a = (-kx-bv)/m, to replace the first term in the above equation. This will give us:

dE/dt = m((-kx-bv)/m)(v) + (-kx-bv)

Simplifying, we get:

dE/dt = (-kx-bv)(v) + (-kx-bv)

Expanding this, we get:

dE/dt = -kv^2-bvv - kxv-bv

Simplifying further, we finally get:

dE/dt = -bv^2

This shows that the time rate of change of mechanical energy for a damped, undriven oscillator is given by -bv^2, which is always negative. This makes sense, as a damped oscillator will lose energy over time due to friction and other dissipative forces, resulting in a decrease in
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top