Solving the Initial Value Problem for Impulse Response with Peak Value of 2

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SUMMARY

The discussion focuses on solving the initial value problem defined by the differential equation y'' + (gamma)y' + y = k(delta)(t-1) with initial conditions y(0) = 0 and y'(0) = 0, where (gamma) is set to 1/2. The goal is to determine the value of k, denoted as k1, that results in a peak response value of 2. The solution involves using Laplace transforms and the method of undetermined coefficients, leading to the formula k1 = (5/2)/(delta(ln(4/3) - 1) for the specified peak value.

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  • Knowledge of the method of undetermined coefficients for finding particular solutions.
  • Basic concepts of damped harmonic oscillators and impulse response analysis.
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huk
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hey,

Consider the initial value problem:

y''+(gamma)y'+y=k(delta)(t-1), y(0)=0, y'(0)=0

where k is the magnitude of an impulse at t = 1 adn (gamma) is the damping coefficietnt(for resistance).
Let (gamma)=1/2. find the value of k for which the response has a peak value of 2; call this value k1..

evnthough I am taking this diff equ course, i have difficult understanding the physical meaning of this equation... what does peak value mean... and how do i find k w/ this peak value?

thanks a lot
 
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The response of a system is the solution to the initial value problem. What you want to do is solve the initial value problem via Laplace transforms...your y(t) will be a function of both time and k, set y(1) = 2 and solve for k
 


To solve for the value of k1, we first need to understand the physical meaning of the equation. The initial value problem represents a damped harmonic oscillator with an impulse applied at t=1. The damping coefficient (gamma) represents the resistance in the system, and the impulse (k) represents the magnitude of the force applied. The peak value refers to the maximum amplitude of the response of the oscillator.

To find k1, we can use the method of undetermined coefficients. We assume a solution of the form y = Aet, where A is a constant to be determined. Substituting this into the equation, we get:

Ae^(t/2) + (gamma)Ae^(t/2) + Ae^(t/2) = k(delta)(t-1)

Simplifying, we get:

2Ae^(t/2) + (gamma)Ae^(t/2) = k(delta)(t-1)

At the peak value, the derivative of the solution is equal to 0. So we can set y' = 0 and solve for t to find the time at which the peak occurs. This gives us t = ln(2/(gamma+2)). Substituting this value into the equation, we get:

2Ae^(ln(2/(gamma+2))/2) + (gamma)Ae^(ln(2/(gamma+2))/2) = k(delta)(ln(2/(gamma+2)) - 1)

Simplifying, we get:

A(2 + (gamma)) = k(delta)(ln(2/(gamma+2)) - 1)

To find k1, we want the peak value to be 2. So we set A = 1 and solve for k:

2 + (gamma) = k(delta)(ln(2/(gamma+2)) - 1)

k = (2 + (gamma))/(delta(ln(2/(gamma+2)) - 1))

Substituting (gamma) = 1/2, we get:

k1 = (5/2)/(delta(ln(4/3) - 1))

Therefore, the value of k for which the response has a peak value of 2 is given by k1 = (5/2)/(delta(ln(4/3) - 1)). I hope this helps clarify the physical meaning of the equation and how to find the value of k for
 

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