Question

A block oscillating on a spring has an amplitude of 20 cm. What will the amplitude be if the total energy is doubled? Explain.

Answer 1

Answer:28.28 cm

Explanation:

Given

Amplitude $A=20\ cm$

Total energy of system is equal to the Elastic Potential Energy of the spring mass system at maximum amplitude which is given by

$E=U=\frac{1}{2}kA^2$

$E=\frac{1}{2}kA^2$

When energy is doubled then

$2E=\frac{1}{2}kA'^2$

$2\times \frac{1}{2}kA^2= \frac{1}{2}kA'^2$

$A'=\sqrt{2}A$

$A'=1.414\times 20$

$A'=28.28\ cm$