Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Answer:
The spring constant is approximately 46.382 newtons per meter.
Explanation:
From Physics, the period ([tex]T[/tex]), measured in seconds, experimented by an object under Simple Harmonic Motion:
[tex]T = 2\pi\cdot \sqrt{\frac{W}{g\cdot k} }[/tex] (1)
Where:
[tex]W[/tex] - Weight, measured in newtons.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
If we know that [tex]W = 127\,N[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]T = 3.32\,s[/tex], then the spring constant of the system is:
[tex]\frac{T^{2}}{4\cdot \pi^{2}} = \frac{W}{g\cdot k}[/tex]
[tex]k = \frac{4\cdot \pi^{2}\cdot W}{g\cdot T^{2}}[/tex]
[tex]k \approx 46.382\,\frac{N}{m}[/tex]
The spring constant is approximately 46.382 newtons per meter.
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.