State the initial value problem describing the mass's motion upon starting from its release, Solution: Displacement y from equilibrium solves the equation given in the lecture y" the initial values given in the problem are y(0) = 0 and y'(0) = 0.01. Solve the initial value problem formulated in (b). Solution: The root oquation has solutions I k ===y=-500y, m 500 So the general coltuion is
State the initial value problem describing the mass's motion upon starting from its release, Solution: Displacement y from equilibrium solves the equation given in the lecture y" the initial values given in the problem are y(0) = 0 and y'(0) = 0.01. Solve the initial value problem formulated in (b). Solution: The root oquation has solutions I k ===y=-500y, m 500 So the general coltuion is
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can you explain step by step how to do PART C pls
![**Problem 2:** A 100 gram mass, when attached to a spring hanging vertically, stretches the spring 2 cm. The mass is then released from the equilibrium position with an initial velocity of 1 cm/s. Assume that there is no damping.
**(a) Determine the spring constant \( k \),**
**Solution:** Equating the gravitational force and spring force, we get
\[
k = \frac{mg}{\Delta x} = \frac{10 \times 0.1}{0.02} = 50 \, \text{N/m}.
\]
**(b) State the initial value problem describing the mass’s motion upon starting from its release,**
**Solution:** Displacement \( y \) from equilibrium solves the equation given in the lecture
\[
y'' = -\frac{k}{m}y = -500y,
\]
the initial values given in the problem are \( y(0) = 0 \) and \( y'(0) = 0.01 \).
**(c) Solve the initial value problem formulated in (b).**
**Solution:** The root equation has solutions \( \lambda = \pm \sqrt{500}i \). So the general solution is
\[
y(t) = C_1 \sin(\sqrt{500}t) + C_2 \cos(\sqrt{500}t).
\]
Using the initial values, I get \( C_2 = 0 \) and \( \sqrt{500}C_1 = 0.01 \), so
\[
y(t) = \frac{0.01}{\sqrt{500}} \sin(\sqrt{500}t).
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faeefd442-7c64-4a51-8ed1-c0196ac6a13e%2F7907857c-34e3-4b62-b7ae-8932f98c4e9e%2Fafbllgd_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 2:** A 100 gram mass, when attached to a spring hanging vertically, stretches the spring 2 cm. The mass is then released from the equilibrium position with an initial velocity of 1 cm/s. Assume that there is no damping.
**(a) Determine the spring constant \( k \),**
**Solution:** Equating the gravitational force and spring force, we get
\[
k = \frac{mg}{\Delta x} = \frac{10 \times 0.1}{0.02} = 50 \, \text{N/m}.
\]
**(b) State the initial value problem describing the mass’s motion upon starting from its release,**
**Solution:** Displacement \( y \) from equilibrium solves the equation given in the lecture
\[
y'' = -\frac{k}{m}y = -500y,
\]
the initial values given in the problem are \( y(0) = 0 \) and \( y'(0) = 0.01 \).
**(c) Solve the initial value problem formulated in (b).**
**Solution:** The root equation has solutions \( \lambda = \pm \sqrt{500}i \). So the general solution is
\[
y(t) = C_1 \sin(\sqrt{500}t) + C_2 \cos(\sqrt{500}t).
\]
Using the initial values, I get \( C_2 = 0 \) and \( \sqrt{500}C_1 = 0.01 \), so
\[
y(t) = \frac{0.01}{\sqrt{500}} \sin(\sqrt{500}t).
\]
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