Question

Show that the total energy of a stretched spring remains, conserved when it is released, find the expression for the maximum speed.

Answer 1

Spring force $=-kx$

$\Delta$ (work done) $=F.dx$

Total work done $=\int F.dx$

$\int (-kx)dx$

${\int }_{x_2}^{x_1}-kxdx$

$w=\frac{1}{2}k{x}_1^2-\frac{1}{2}k{x}_2^2$

Work done =change in kinetic energy

$=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2$

$\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2$$=\frac{1}{2}k{x}_1^2-\frac{1}{2}k{x}_2^2$

$\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2=C$

Maximum velocity $=\sqrt{\frac{k}{m}}.x$