Oscillations: Question 5

Worked solution

Step 1: Recall the velocity-displacement equation for SHM

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

Step 2: Find ω from the period

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.50}=12.566\text{ rad s}^{-1}$

Step 3: Substitute the displacement

$x_0^2-x^2=(0.12)^2-(0.072)^2=0.0144-0.005184=0.009216$ $\sqrt{0.009216}=0.0960\text{ m}$ $v=12.566\times0.0960=1.206\text{ m s}^{-1}$

Recompute as a check, factoring differently: since $x=0.072=0.6\times0.12=0.6x_0$, we have $x_0^2-x^2=x_0^2(1-0.6^2)=x_0^2(1-0.36)=0.64x_0^2$, so $\sqrt{x_0^2-x^2}=0.8x_0=0.8\times0.12=0.096\text{ m}$, the same value as Step 3. Then $v=12.566\times0.096=1.206\text{ m s}^{-1}$, confirming the result.

Rounding to 3 significant figures, $v\approx\boxed{1.21}\text{ m s}^{-1}$.

Why the other options are wrong

• B ($1.51\text{ m s}^{-1}$): this is the maximum speed $v_{max}=\omega x_0=12.566\times0.12=1.508\text{ m s}^{-1}$, which only occurs at $x=0$, not at $x=0.072\text{ m}$.
• C ($0.90\text{ m s}^{-1}$): comes from using $v=\omega x$ instead of $v=\omega\sqrt{x_0^2-x^2}$: $12.566\times0.072=0.905\text{ m s}^{-1}$, this substitutes the current displacement into the wrong (maximum-speed) formula.
• D ($0.60\text{ m s}^{-1}$): comes from using $v=\omega(x_0-x)$ instead of the square-root relationship: $12.566\times(0.12-0.072)=12.566\times0.048=0.603\text{ m s}^{-1}$.

Final answer

• The speed of the particle at $x=0.072\text{ m}$ is $\boxed{1.21}\text{ m s}^{-1}$, option A.