a) Determine Solution: the spring constant k, Equating the gravitational fore
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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Where does the 10 and 0.01 come from
![**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 1 cm/s. Assume that there is no damping.
(a) **Determine the spring constant \( k \),**
**Solution:** Equating the gravitational force and spring force I get
\[
k = \frac{mg}{\Delta x} = \frac{10 \times 0.1}{0.02} = 50 \, \text{Nm}^{-1}
\]
**Explanation:**
- The problem involves calculating the spring constant (\( k \)) of a vertically hanging spring with a mass attached to it.
- The formula used is derived from balancing the gravitational force (\( mg \)) and the spring force (\( k \Delta x \)), where:
- \( m = 0.1 \) kg (100 grams converted to kilograms)
- \( g = 10 \, \text{m/s}^2 \) (assumed acceleration due to gravity)
- \( \Delta x = 0.02 \) m (2 cm converted to meters)
- The calculation shows that the spring constant \( k = 50 \, \text{Nm}^{-1} \), which indicates the stiffness of the spring.
There are no graphs or diagrams in this image.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faeefd442-7c64-4a51-8ed1-c0196ac6a13e%2F9125ce11-47fd-4ce2-ac9a-0ede62ffff66%2F5z1t3xb_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 1 cm/s. Assume that there is no damping.
(a) **Determine the spring constant \( k \),**
**Solution:** Equating the gravitational force and spring force I get
\[
k = \frac{mg}{\Delta x} = \frac{10 \times 0.1}{0.02} = 50 \, \text{Nm}^{-1}
\]
**Explanation:**
- The problem involves calculating the spring constant (\( k \)) of a vertically hanging spring with a mass attached to it.
- The formula used is derived from balancing the gravitational force (\( mg \)) and the spring force (\( k \Delta x \)), where:
- \( m = 0.1 \) kg (100 grams converted to kilograms)
- \( g = 10 \, \text{m/s}^2 \) (assumed acceleration due to gravity)
- \( \Delta x = 0.02 \) m (2 cm converted to meters)
- The calculation shows that the spring constant \( k = 50 \, \text{Nm}^{-1} \), which indicates the stiffness of the spring.
There are no graphs or diagrams in this image.
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