Spring constant k = 27.2128 N/m

?The stretching distance d = 0.077 m

Step-by-step explanation

Given the following data

Mass = 0.213-kg

Stretching distance d

oscillations = 102

Time = 56.7 s. 

Find the A) spring constant

Solution

Formula

T = 2π√(m/k)

Where T is the time period

T = Time/oscillations

T = 56.7s/102

m is the Mass = 0.213-kg

k is the spring constant

T = 2π√(m/k)

56.7s/102 = 2π√(0.213-kg/k)

((56.7s/102)/(2π))² = 0.213-kg/k

k = (0.213-kg)/((56.7s/102)/(2π))²

k = 27.2128 N/m

Spring constant k = 27.2128 N/m

 

B) the stretch distance, d. Complete A and B.

Solution

Force = kx

but also Force = mg

kx = mg

x = mg/k

Where x is the stretching distance = d

m is the Mass = 0.213-kg

g is the gravity = 9.8 m/s²

k is the spring constant k = 27.2128 N/m

x = (0.213-kg)(9.8 m/s²)/(27.2128 N/m)

x = 0.077 m

The stretching distance d = 0.077 m