question_answer

If stretch in a spring of force constant k is doubled, calculate (a) ratio of final to initial force in the spring.         (b) ratio of elastic energies stored in the two cases. (c) work done in changing to the state of double stretch.                

Answer:

                (a) For a given spring, \[\text{F}=\text{kx}\]                       \[\therefore \] \[\frac{{{F}_{2}}}{{{F}_{1}}}=\frac{k{{x}_{2}}}{k{{x}_{1}}}=\frac{2x}{x}=2\] (b) For a given spring, \[\text{U}=\frac{1}{2}\text{k}{{\text{x}}^{\text{2}}}\]     \ \[\frac{{{U}_{2}}}{{{U}_{1}}}=\frac{\frac{1}{2}kx_{2}^{2}}{\frac{1}{2}kx_{1}^{2}}=\frac{{{(2x)}^{2}}}{{{x}^{2}}}=4\] (c) As work done in stretching the spring is stored in the spring in the form of elastic potential energy of the spring, therefore, \[\text{W}={{\text{U}}_{\text{2}}}-{{\text{U}}_{\text{1}}}=\frac{1}{2}kx_{2}^{2}-\frac{1}{2}kx_{1}^{2}=\frac{1}{2}k[{{(2x)}^{2}}-{{x}^{2}}]=\frac{3}{2}k{{x}^{2}}\]