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Lecture

# L20_energy.pdf

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School
University of Toronto St. George
Department
Physics
Course
PHY354H1
Professor
Erich Poppitz
Semester
Winter

Description
Lecture 20: Energy and Oscillations ▯ relevant text sections: Morin 5.1-5.2 Conservation laws ▯ Conservation laws are a central part of modern physics, both for prob- lem solving and for physical insights. ▯ It is not the case that you need a conservation law to solve a system, but often a conservation law gives you a shortcut to solving a problem that helps you avoid the use of F = ma or other even more complicated methods. Conservation of energy in one dimension ▯ Consider one dimensional motion under a force that is a function of x ONLY (constant forces okay too). In that case we can use our sneaky substitution trick from Chapter 3 to remove time from the problem: We have F(x) = ma = m dv = m dv dx = mv dv (1) dt dx dt dx ▯ Keep in mind that the velocity v = v(x). Solving by separation of variables: mvdv = F(x)dx Z v Z x 0 0 mvdv = F(x )dx v0 Zxx 1 2 1 2 0 0 mv ▯ mv 0 = F(x )dx 2 2 x0 ▯ Now we de▯ne a function V (x) such that dV F(x) = ▯ ) dx 1 we can ▯nd V(x) for any F(x) using: Z x 0 0 V (x) = ▯ F(x )dx + V 0 with V (x 0 = V 0 (2) x0 I can then isolate for the integral of F in this equation and plug it into my energy equation: 1 2 1 2 mv ▯ mv = V 0 V (x0 2 2 ▯ If we put all the 0 stu▯ on one side and the other stu▯ on the other side we get: 1 2 1 2 mv 0 V =0 mv + V (x): 2 2 ▯ Since the left hand side is a constant, and the right hand side is true for an arbitrary x, it must apply for ANY x. This means that the RHS of the equation is always equal to a constant, it is conserved. ▯ We de▯ne the mechanical energy as: E = 1 mv + V (x) = K(x _) + V (x): 2 where V (x) is the potential energy and K = mv is the kinetic energy. 2 ▯ We found above that the mechanical energy is ALWAYS constant when you have a F = F(x). Thus we call F(x) a conservative force, because objects moving in F maintain constant energy. NOTE: You ONLY de▯ne a potential energy for forces that are conservative. ▯ We can ▯nd an explicit solution for any energy conserving system: since 1 2 mv = E ▯ V (x) 2 (remember E is a constant) then r v(x) = ▯ 2 [E ▯ V (x)] = dx m dt 2 ▯ This is a ▯rst order ODE, unlike Newton’s law, which is second order! (Of course, this is because we already integrated once to get the energy equation). ▯ Using separation of variables, we can ▯nd: Z x Z x dx 0 dx0 t ▯ 0 = ▯ 0 = ▯ q : x0v(x ) x0 2 [E ▯ V (x )] m From this we can obtain t = t(x). Also, if the expression is not too gruesome, we can invert it to ▯nd x(t). Potential Energy of a Spring ▯ If F(x) = ▯kx; where x is the departure of the spring from equilibrium, 1 2 then using equation (2) we ▯nd V (x) = 2kx and the mechanical energy is 1 2 1 2 E = mx_ + kx : 2 2 ▯ If the spring force is the only force acting, then we have simple harmonic motion with frequency ! = k=m (old news by now). Notice that this is actually the ratio of the coe▯cients of the KE and PE terms. In fact, we can write 1 2 2 2 E = m(x_ + ! x ): 2 ▯ In general, if we can write the energy equation in the form 1 1 E = a_ + bq 2 (3) 2 2 where q is some variable related to position and E is some constant, 2 we know that the frequency will satisfy ! = b=a. If you want proof, you just need to work with the F = ma equation and our method for ▯t solving by trying solutions Ae and solving for ▯. All simple harmonic oscillators (SHOs) have quadratic potentials. 3 Potential energy of a small angle pendulum ▯ Okay, based on what we learned from the spring, we should be able to ▯nd the frequency of the pendulum by ▯nding the potential energy, writing out the mechanical energy equation and taking the ratio of the coe▯cients in front of the position and velocity terms. ▯ Lets write the position in terms of angle. For the kinetic energy, we need to di▯erentiate the position: x = l▯ ) x _ = l▯ So the kinetic energy is K = 1mv = m(l▯) . _2 2 2 ▯ Setting the potential energy to zero when the pendulum is at ▯ = 0, we get the potential energy by integrating the force (gravity in this case): Z h V = mgdx = mgh = mgl(1 ▯ cos▯) 0 For the small angle approximation, if we use a Taylor expansion of 2 cos▯ ▯ 1 ▯ ▯ =2 we get: V (▯) ▯ 1mgl▯ : 2 So the mechanical energy is E = 1ml ▯ + mgl(1 ▯ cos▯) ▯ 1ml ▯ + m
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