The potential energy of a simple harmonic oscillator when the...

The potential energy of a simple harmonic oscillator at a displacement $x$ is given by $U = \dfrac{1}{2} k x^2$.

The total energy of the oscillator is $E = \dfrac{1}{2} k A^2$.

Given the displacement $x = \dfrac{3}{4}A$, substituting this into the potential energy formula gives:

$U = \dfrac{1}{2} k \left(\dfrac{3}{4}A\right)^2$

$U = \dfrac{9}{16} \left(\dfrac{1}{2} k A^2\right)$

Since $E = \dfrac{1}{2} k A^2$, we get:

$U = \dfrac{9}{16}E$

Answer: $\dfrac{9}{16}E$