A mass of 1 kg is hung vertically from a spring stretching the spring .20 m. It

Question

A mass of 1 kg is hung vertically from a spring stretching the spring .20 m. It is then pulled .10 m below its equilibrium position and released from rest. Find

(a) the angular frequency of oscillation

(b) the spring constant of the spring

(c) the amplitude of oscillation

(d) the kinetic energy of mass when passing through equilibrium

(e) the potential energy of the spring when the mass passes through equilibrium

(f) the time to first reach equilibrium

(g) the time to make one complete cycle.

Please explain and answer thouroughly.

Thanks!

Explanation / Answer

a) elongation in spring at equilibrium=0.2m

at equilibriium, mg=k*0.2 is

k=mg/x=(1*9.8)/0.2=49 N/m

angular frequency of oscillation= (k/m)^0.5=(49/1)^0.5=7 rad/seconf

b)spring constant of the spring= 49 N/m

c)amplitude of oscillation=0.1m

d)kinetic energy of mass when passing through equilibrium= mg*0.1=0.98J

e)potential energy of the spring when the mass passes through equilibrium=(1/2)*k*0.2^2=0.98J

f)the time to first reach equilibrium=time period/4=2*0.3.14/4*7=0.224second

g) time period=2*3.14/7=0.8971second

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