Oscillations: Question 8

Worked solution

Part (a): Show that ω = 9.49 rad s⁻¹

For a mass on a spring, the angular frequency is: $\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{18}{0.200}}=\sqrt{90}$

Recompute as a check, evaluating the division first in a different way: $18\div0.200=90.0$ exactly, and $\sqrt{90}=9.4868$.

So $\omega=9.49\text{ rad s}^{-1}$ (3 s.f.), as required to show. This exact value $\omega^2=90\text{ rad}^2\text{ s}^{-2}$ is carried forward for accuracy.

Part (b): Total energy

The total (constant) energy of an SHM oscillation is: $E=\tfrac12m\omega^2x_0^2$

Substituting $m=0.200\text{ kg}$, $\omega^2=90\text{ rad}^2\text{ s}^{-2}$, $x_0=0.050\text{ m}$: $E=\tfrac12\times0.200\times90\times(0.050)^2=0.100\times90\times0.0025$

Working step by step: $0.100\times90=9.00$; then $9.00\times0.0025=0.0225\text{ J}$.

Recompute in a different grouping as a check: $(0.050)^2\times90=0.0025\times90=0.225$; then $0.100\times0.225=0.0225\text{ J}$, the same result.

So $E=\boxed{2.25\times10^{-2}}\text{ J}$, i.e. $22.5\text{ mJ}$.

Part (c): Show that x = 2.89 × 10⁻² m when KE = 2PE

At any displacement, $KE+PE=E$ (total energy is constant). If $KE=2PE$, then: $2PE+PE=E \implies 3PE=E \implies PE=\frac{E}{3}$

Since $PE=\tfrac12m\omega^2x^2$ and $E=\tfrac12m\omega^2x_0^2$: $\tfrac12m\omega^2x^2=\frac{\tfrac12m\omega^2x_0^2}{3} \implies x^2=\frac{x_0^2}{3} \implies x=\frac{x_0}{\sqrt3}$

Substituting $x_0=0.050\text{ m}$: $x=\frac{0.050}{\sqrt3}=\frac{0.050}{1.7321}=0.028868\text{ m}$

Recompute as a check, working directly in energy terms: $PE=E/3=0.0225/3=0.00750\text{ J}$. Then $x^2=\dfrac{2PE}{m\omega^2}=\dfrac{2\times0.00750}{0.200\times90}=\dfrac{0.0150}{18.0}=8.333\times10^{-4}\text{ m}^2$, so $x=\sqrt{8.333\times10^{-4}}=0.028868\text{ m}$, the same value, confirming the result.

So $x=\boxed{2.89\times10^{-2}}\text{ m}$ (3 s.f.), as required to show.

Part (d): Speed at this displacement

$v=\omega\sqrt{x_0^2-x^2}$

Substituting $x_0^2=(0.050)^2=0.0025\text{ m}^2$ and $x^2=x_0^2/3=8.333\times10^{-4}\text{ m}^2$: $x_0^2-x^2=0.0025-0.0008333=0.0016667\text{ m}^2$ $\sqrt{0.0016667}=0.040825\text{ m}$ $v=9.4868\times0.040825=0.38730\text{ m s}^{-1}$

Recompute as a check, using $KE=\tfrac12mv^2$ directly: since $KE=2E/3=2\times0.0225/3=0.0150\text{ J}$, then $v^2=\dfrac{2\times KE}{m}=\dfrac{2\times0.0150}{0.200}=0.150\text{ m}^2\text{s}^{-2}$, so $v=\sqrt{0.150}=0.3873\text{ m s}^{-1}$, the same value, confirming the result.

So $v=\boxed{0.387}\text{ m s}^{-1}$ (3 s.f.).

Final answers

• (a) $\omega=\boxed{9.49}\text{ rad s}^{-1}$
• (b) $E=\boxed{2.25\times10^{-2}}\text{ J}$
• (c) $x=\boxed{2.89\times10^{-2}}\text{ m}$
• (d) $v=\boxed{0.387}\text{ m s}^{-1}$