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A spring of spring constant 30.0 N/m is attached to a 2.3 kg mass and set in motion. What is the period and frequency of vibration for the 2.3 kg mass?

Sagot :

Eduard22slyidn

Answer:

1. The period is 1.74 s.

2. The frequency is 0.57 Hz

Explanation:

1. Determination of the the period.

Spring constant (K) = 30 N/m

Mass (m) = 2.3 Kg

Pi (π) = 3.14

Period (T) =?

The period of the vibration can be obtained as follow:

T = 2π√(m/K)

T = 2 × 3.14 × √(2.3 / 30)

T = 6.28 × √(2.3 / 30)

T = 1.74 s

Thus, the period of the vibration is 1.74 s.

2. Determination of the frequency.

Period (T) = 1.74 s

Frequency (f) =?

The frequency of the vibration can be obtained as follow:

f = 1/T

f = 1/1.74

f = 0.57 Hz

Thus, the frequency of the vibration is 0.57 Hz

Pstnonsonjokuidn

The period of the vibration is  1.76 s and the frequency of the vibration is  0.57 s-1.

Using the formula;

T = 2π√(m/K)

Where;

T = period

m = mass

K = spring constant

Substituting values;

T = 2(3.142)√2.3/30

T = 6.284 × 0.28

T = 1.76 s

Recall that the period is the inverse of frequency;

f = 1/T

f = 1/1.76 s

f = 0.57 s-1

Learn more about spring constant: https://brainly.com/question/4291098

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