Question

A block oscillating on a spring has an amplitude of 20 cmcm. Part A Part complete What will the block's amplitude be if its total energy is doubled

Answer 1

Hi there!

Recall that the amplitude of an oscillating system occurs at a point of maximum potential energy.

For a spring system, potential energy is given as:
$U = \frac{1}{2}kx^2$

U = Potential Energy (J)

k = Spring constant (N/m)
x = amplitude (m)

We can rearrange the equation to solve for amplitude more easily.

$U = \frac{1}{2}kx^2\\\\2U = kx^2\\\\x^2 = \frac{2U}{k}\\\\x = \sqrt{\frac{2U}{k}}$

If we double 'U':
$x' = \sqrt{\frac{2(2U)}{k}} = \sqrt{2}*\sqrt{\frac{2U}{k}}\\\\x' = \sqrt{2} * x$

Thus, the new amplitude would be √2 times greater, so:
$20 cm \cdot \sqrt{2} = \boxed{28.284 cm}$