Question

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.

Answer 1

The spring constant is approximately 46.382 newtons per meter.

Step-by-step explanation:

From Physics, the period (
`T`), measured in seconds, experimented by an object under Simple Harmonic Motion:

`T = 2\pi\cdot \sqrt{(W)/(g\cdot k) }` (1)

Where:

`W` - Weight, measured in newtons.

`g` - Gravitational acceleration, measured in meters per square second.

`k` - Spring constant, measured in newtons per meter.

If we know that
`W = 127\,N`,
`g = 9.807\,(m)/(s^(2))` and
`T = 3.32\,s`, then the spring constant of the system is:

`(T^(2))/(4\cdot \pi^(2)) = (W)/(g\cdot k)`

`k = (4\cdot \pi^(2)\cdot W)/(g\cdot T^(2))`

`k \approx 46.382\,(N)/(m)`

The spring constant is approximately 46.382 newtons per meter.