a 0.40kg mass is attached to an ideal spring. if it oscillates horizontally with a period of 0.50s and an amplitude of 0.20m, then determine the spring constant of the spring? group of answer choices

Respuesta :

By angular speed, the spring constant is 63.21 N/m.

We need to know about the angular speed of harmonic oscillation. the angular speed can be determined as

ω² = k / m

where ω is angular speed, k is spring constant,  and m is the mass.

The angular speed can be calculated by

ω = 2π/T

Hence,

ω² = k / m

(2π/T)² = k / m

where T is period.

From the question above, we know that:

m = 0.4 kg

T = 0.5 s

By substituting to the equation, we get

(2π/T)² = k / m

(2π/0.5)² = k / 0.4

158.04 = k / 0.4

k = 63.21 N/m

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