A 127 N object vibrates with a period of 3.32 s when hanging from a spring. What is the spring constant of the spring? The acceleration of gravity is 9.81 m/s 2 . Answer in units of N/m.

Respuesta :

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.