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 = (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 = (1)/(2)kx^2\\\\2U = kx^2\\\\x^2 = (2U)/(k)\\\\x = \sqrt{(2U)/(k)}`

If we double 'U':

`x' = \sqrt{(2(2U))/(k)} = √(2)*\sqrt{(2U)/(k)}\\\\x' = √(2) * x`

Thus, the new amplitude would be √2 times greater, so:

`20 cm \cdot √(2) = \boxed{28.284 cm}`