Question

When a force of 20.0 N is applied to a spring, it elongates 0.20 m. Determine the period of oscillation of a 4.0-kg object suspended from this spring.

Answer 1

Answer:

1.26 secs.

Explanation:

The following data were obtained from the question:

Force (F) = 20 N

Extention (e) = 0.2 m

Mass (m) = 4 Kg

Period (T) =.?

Next, we shall determine the spring constant, K for spring.

The spring constant, K can be obtained as follow:

Force (F) = 20 N

Extention (e) = 0.2 m

Spring constant (K) =..?

F = Ke

20 = K x 0.2

Divide both side by 0.2

K = 20/0.2

K = 100 N/m

Finally, we shall determine the period of oscillation of the 4 kg object suspended on the spring. This can be achieved as follow:

Mass (m) = 4 Kg

Spring constant (K) = 100 N/m

Period (T) =..?

T = 2π√(m/K)

T = 2π√(4/100)

T = 2π x √(0.04)

T = 2π x 0.2

T = 1.26 secs.

Therefore, the period of oscillation of the 4 kg object suspended on the spring is 1.26 secs.