Question

A suture is stretched to 10%. When the stress is released it recovered 35 % of its strain after 5 hr. What is the retardation time in hours. What is the amount of strain recovered (in %) after 6 hrs of its original value.

Answer 1

$$\frac{l}{l_0}(t)=1-e^{-\frac{t}{\tau}}$$

1)

$$\frac{l}{l_0}(5)=1-e^{-\frac{5}{\tau}}=0.35$$

$$\tau=11.6\ h$$

2)

$$\frac{l}{l_0}(6)=1-e^{-\frac{6}{11.6}}=0.40$$

The amount of strain recovered after 6 hrs is 40% of its original value.