Question

Two springs of spring constants $1500N/m$ and $3000N/m$ respectively are stretched with the same force. They will have the potential energies in the ratio:

• A. $1:2$
  
• B. $2:1$
  
• C. $1:4$
  
• D. $4:1$
  

Answer 1

The correct option is B $2:1$

When spring are streached to same distance,x is same

by formula$PE=\frac{1}{2}\ast k\ast x^2$

therefore $\frac{P{E}_1}{P{E}_2}=\frac{k_1}{k_2}=\frac{3000}{1500}=2:1$