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A 125 N object vibrates on a spring with a spring constant of 39.6 N/m. What is the period of the vibration?

User Thrusty
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1 Answer

3 votes

Answer:

3.53 s

Step-by-step explanation:

From the question given above, the following data were obtained:

Force (F) = 125 N

Spring constant (K) = 39.6 N/m

Period (T) =?

Next, we shall determine the mass of the object. This can be obtained as follow:

Force (F) = 125 N

Acceleration due to gravity (g) = 10 m/s²

Mass (m) =?

F = m × g

125 = m × 10

Divide both side by 10

m = 125 / 10

m = 12.5 Kg

Finally, we shall determine the period of vibration. This can be obtained as follow:

Mass (m) = 12.5 Kg

Spring constant (K) = 39.6 N/m

Pi (π) = 3.14

Period (T) =?

T = 2π√(m/K)

T = 2 × 3.14 × √(12.5/39.6)

T = 6.28 × √(12.5/39.6)

T = 3.53 s

Thus, the period of vibration is 3.53 s

User Tylor
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