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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 173
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

Vibrating Spring without Damping. A vibrating spring without damping can be modeled by the initial value problem (11) in Example 3 by taking b=0.

a) If m=10 kg,k=250 kg/sec2,y(0)=0.3 m, and y'(0)=-0.1 m/sec, find the equation of motion for this undamped vibrating spring.

b) After how many seconds will the mass in part (a) first cross the equilibrium point?

c) When the equation of motion is of the form displayed in (9), the motion is said to be oscillatory with frequency β/2π. Find the frequency of oscillation for the spring system of part (a).

  1. The equation of motion for vibrating spring is y(t)=0.3cos(5t)-0.02sin(5t)
  2. The mass crosses the equilibrium at t=0.3 seconds
  3. The frequency of the spring is f=52π
See the step by step solution

Step by Step Solution

Step 1: Differentiating the equation.

The differential equation without damping is my''+ky=0

Given m=10 kg and k=250 kg/sec2

Let y=ert

Then y'(t)=rert


Then the auxiliary equation is 10r2+250=0


Therefore, the general solution is y(t)=c1cos(5t)+c2sin(5t).

Step 2: Finding c1 and  c2

Given initial conditions are y(0)=0.3 and y'(0)=-0.1


y(0)=c1cos(0)+c2sin(0) c1=0.3

And y'(t)=-5c1sin(5t)+5c2cos(5t)

y'(0)=-5c1sin(0)+5c2cos(0) 5c2=-0.1 c2=-0.02

Therefore, the solution is y(t)=0.3cos(5t)-0.02sin(5t) .

Step 3: Finding the time.

When the spring crosses the equilibrium y(t)=0 , so we have to find the t

0.3cos(5t)-0.02sin(5t)=0 0.3cos(5t)=0.02sin(5t) tan(5t)=0.30.02 5t=arctan(15) 5t1.504t0.3

So, at t=0.3 seconds the mass crosses the equilibrium

Step 4: Finding the frequency

Here β=5, the Frequency of the spring is f=52π.

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