Mastering Oscillations: Formulas, Traps & Exam Strategy | ExamiFly

The maximum speed in SHM is given by $v_{max} = A\omega$. The general expression for speed at displacement $x$ is $v = \omega \sqrt{A^2 - x^2}$.

We are given that the speed is $v_{max}/2$, so $v = \frac{A\omega}{2}$. Equating this with the general speed formula: $\frac{A\omega}{2} = \omega \sqrt{A^2 - x^2}$.

Cancel $\omega$ from both sides: $\frac{A}{2} = \sqrt{A^2 - x^2}$. Squaring both sides gives $\frac{A^2}{4} = A^2 - x^2$. Rearranging for $x^2$: $x^2 = A^2 - \frac{A^2}{4} = \frac{3A^2}{4}$. Taking the square root, $x = \pm \frac{\sqrt{3}A}{2}$. Thus, the displacement is $\frac{\sqrt{3}A}{2}$.

For SHM, the maximum speed $v_{max}$ occurs at the equilibrium position and is related to the amplitude $A$ and angular frequency $\omega$ by $v_{max} = A\omega$. The angular frequency for a spring-mass system is $\omega = \sqrt{k/m}$.

In the first case, the block is pulled to a distance $x_0$, so the amplitude is $A_1 = x_0$. The maximum speed is given as $v_0 = A_1\omega = x_0 \sqrt{k/m}$.

In the second case, the block is pulled to a distance $2x_0$, so the new amplitude is $A_2 = 2x_0$. The new maximum speed, $v_{max}'$, will be $A_2\omega = (2x_0) \sqrt{k/m}$. Comparing this with $v_0$, we see that $v_{max}' = 2 (x_0 \sqrt{k/m}) = 2v_0$.