The Equation 13-75 mentioned is the following: E(t)= 0.5kA 2 e -bt/m please expl

Question

The Equation 13-75 mentioned is the following:

E(t)= 0.5kA2e-bt/m

please explain your answer. the textbook is confusing me as I read the little they have on this topic.

thanks!!

READ THIS QUESTION CAREFULLY: For non-ideal oscillators (such as a real pendulum) energy is lost and the amplitude (which is the maximum displacement) is no longer constant but also decreases with time. How fast the energy lost is described by a TIME CONSTANT, r = m/b. (Compare this idea to equation 13-75: the exponent must be dimensionless, hence m/b must have the unit of time.) Typically after ONE time constant has elapsed, the system has ...(hint: this is when t = r) LOST half of its initial energy. LOST 13% of its initial energy. lost NO energy. LOST 37% of its initial energy. LOST 63% of its initial energy.

Explanation / Answer

lost 63 % of energy,

dE/dt =( 0.5 * K*A^2 * ( -b/m) e^(-bt/m) )

at t=0 , dE/dt = 0.5 * K*A^2 * (-b/m) * e^(0) = 0.5 * K*A^2 * (-b/m)

so time constant is defined as the time required for the energy to become zero when the energy decreases at the rate of (dE/dt) at t=0

so, E(t =0) + ( tc) * ( (dE/dt) at t=0) = 0

0.5*K*A^2 = tc * 0.5 * K*A^2 * (b/m)

tc = m/b

so, at t=tc, E (t =tc) = 0.5 *K*A^2 *e^(-b*m/b*m) = 0.5 *K*A^2 *e^(-1) = 0.3678 *0.5 *K*A^2

so, E(t=0) - E (t =tc)/ E(t=0) = 0.5 *K*A^2 - 0.3678 *0.5 *K*A^2 / 0.5 *K*A^2 = 0.6322*0.5 *K*A^2/0.5 *K*A^2 = 0.6322

In % , it is 63.22 %

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