Energy transformations in oscillations (HL)

Answer

$E_T = \tfrac{1}{2}m\omega^2 x_0^2$ — a **constant**, set by the amplitude. Unit: **J**.

Answer

$E_K = \tfrac{1}{2}m\omega^2(x_0^2 - x^2)$ — biggest at the centre. Unit: **J**.

Answer

$E_P = E_T - E_K = \tfrac{1}{2}m\omega^2 x^2$ — biggest at the extremes. Unit: **J**.

Answer

$E_T \propto x_0^2$ — **double the amplitude → four times** the energy.

Answer

$E_T \propto \omega^2$ — **double ω → four times** the energy.

Answer

At the **centre** (x = 0), where it equals the total energy E_T.

Answer

At the **extremes** (x = ±x₀), where it equals the total energy E_T.

Answer

The **total energy** E_T. Kinetic and potential just **interchange**: E_K + E_P = E_T always.

Answer

Subtract: $E_P = E_T - E_K$ — don't recompute from scratch.

Answer

Both formulas use **x²**, so x = +0.06 m and x = −0.06 m give the **same** energies.

Answer

$E_T = \tfrac{1}{2}(0.20)(25)(0.010) = 0.025$ J.

Answer

Use $E_K = \tfrac{1}{2}mv^2$ and solve for $v = \sqrt{2E_K/m}$.