A classical mechanics problem by Sachin Vishwakarma

Classical Mechanics Level 3

Let a spring of spring constant $k$ is cut into two parts such that one part is 3 times the other . The spring constant of the longer part will be :

It will remain the same i.e. k

3k

4k

\frac{3}{4}

times of

k

\frac{4}{3}

times of

k

1 solution

Tan LeTian
Mar 3, 2016

Using Hooke's Law, $T= \dfrac{\text{modulusof elasticity}}{\text{natural length}} \times \text{ extension}$

Where Spring constant, $k=\dfrac{\text{modulus of elasticity}}{\text{natural length}}$

Modulus of elasticity is constant, Spring contant is inversely proportional to natural length

Spring contant $\times$ natural length = constant

(K1)( L1) =( K2)(L2) Where K = spring contant of spring L= length of spring

Therefore, Spring constant of longer cut spring $=\frac{4}{3}k$