A mass weighing 9pounds is attached to a spring whose spring constant is 36 lb/ft. What is the period of simple harmonic motion? b. d. a. π 2√2 π 4√2 C. П 3√2 π 6√3 e. π 6√2
A mass weighing 9pounds is attached to a spring whose spring constant is 36 lb/ft. What is the period of simple harmonic motion? b. d. a. π 2√2 π 4√2 C. П 3√2 π 6√3 e. π 6√2
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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![**Problem:**
A mass weighing 9 pounds is attached to a spring whose spring constant is 36 lb/ft. What is the period of simple harmonic motion?
**Answer Choices:**
a. \( \frac{\pi}{2\sqrt{2}} \)
b. \( \frac{\pi}{4\sqrt{2}} \)
c. \( \frac{\pi}{3\sqrt{2}} \)
d. \( \frac{\pi}{6\sqrt{3}} \)
e. \( \frac{\pi}{6\sqrt{2}} \)
**Explanation:**
To find the period (T) of a mass-spring system in simple harmonic motion, we use the formula:
\[
T = 2\pi \sqrt{\frac{m}{k}},
\]
where \(m\) is the mass and \(k\) is the spring constant.
1. First, we need to convert the weight (force) to mass. The weight \(W\) in pounds is given by \(W = mg\), where \(g\) is the acceleration due to gravity (32.2 ft/s² in imperial units). Thus,
\[
m = \frac{W}{g} = \frac{9 \text{ lb}}{32.2 \text{ ft/s}^{2}} \approx 0.2797 \text{ slug}.
\]
2. Now use the given values in the formula for the period \(T\):
\[
T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.2797 \text{ slug}}{36 \text{ lb/ft}}}.
\]
3. Simplify the expression to find the period:
\[
T \approx 2\pi \sqrt{\frac{0.2797}{36}} \approx 2\pi \sqrt{0.007769} \approx 2\pi \times 0.0881 \approx 0.5534 \pi \text{ sec}.
\]
We approximate our result to match the closest provided choices, which addresses \(2\sqrt{2}\) as a connection to the calculation; ensuring all values match the given format.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92c05ba6-e775-4ea8-9e55-a6761b98c61f%2Fdbe87419-b373-4007-993e-e5ad915a3d90%2Fm9qplm8_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
A mass weighing 9 pounds is attached to a spring whose spring constant is 36 lb/ft. What is the period of simple harmonic motion?
**Answer Choices:**
a. \( \frac{\pi}{2\sqrt{2}} \)
b. \( \frac{\pi}{4\sqrt{2}} \)
c. \( \frac{\pi}{3\sqrt{2}} \)
d. \( \frac{\pi}{6\sqrt{3}} \)
e. \( \frac{\pi}{6\sqrt{2}} \)
**Explanation:**
To find the period (T) of a mass-spring system in simple harmonic motion, we use the formula:
\[
T = 2\pi \sqrt{\frac{m}{k}},
\]
where \(m\) is the mass and \(k\) is the spring constant.
1. First, we need to convert the weight (force) to mass. The weight \(W\) in pounds is given by \(W = mg\), where \(g\) is the acceleration due to gravity (32.2 ft/s² in imperial units). Thus,
\[
m = \frac{W}{g} = \frac{9 \text{ lb}}{32.2 \text{ ft/s}^{2}} \approx 0.2797 \text{ slug}.
\]
2. Now use the given values in the formula for the period \(T\):
\[
T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.2797 \text{ slug}}{36 \text{ lb/ft}}}.
\]
3. Simplify the expression to find the period:
\[
T \approx 2\pi \sqrt{\frac{0.2797}{36}} \approx 2\pi \sqrt{0.007769} \approx 2\pi \times 0.0881 \approx 0.5534 \pi \text{ sec}.
\]
We approximate our result to match the closest provided choices, which addresses \(2\sqrt{2}\) as a connection to the calculation; ensuring all values match the given format.
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