question_answer

If a block attached to a spring (whose other end is rigidly fixed) is pulled up to distance \[{{x}_{0}}\] and released, the amplitude of its motion cannot exceed \[\pm {{x}_{0}}\]. Why?

Answer:

                If the amplitude of motion exceeds \[\pm {{x}_{0}},\], then the potential energy would be greater than  \[\frac{1}{2}kx_{0}^{2}\] at \[x>{{x}_{0}}.\]  So its K.E. \[=\frac{1}{2}kx_{0}^{2}-\frac{1}{2}k{{x}^{2}}\] would be negative. This is impossible. Hence the block cannot go beyond \[x=\pm {{x}_{0}}.\]