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?

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

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

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