Aspring-mass system has a spring constant k = 383 n/m, length of the spring l = 0.5 m, and the mass attached to it is m = 3.8 kg. what will be its frequency of oscillation if it is pulled by 5.0 cm from its equilibrium position and released?

Frequency = f = 10.0394 (1/s)

Explanation:

The frequency of oscillation of the system is given by the action:

f= √(k/m)

f= system count

k = spring constant

m = mass connected to the spring

Therefore the frequency will be:

f= √(k/m) = √(383(N/m) / (3.8kg))= √( 100.7895 (kg×m/s²)/(kg ) =

= √( 100.7895 (1/s²) = 10.0394 (1/s)

the answer is a)

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