Source: High school physics (Chinese)
Problem
A tuning fork vibrates in simple harmonic motion at a frequency of $1000$ Hz, and the tip of one of its prongs has an amplitude of $0.40$ mm.
- Find the maximum acceleration and the maximum speed of the prong tip.
- Find the acceleration and the speed when the tip's displacement is $0.20$ mm.
(1) $a_{\max} = 1600\pi^2 \approx 1.6 \times 10^4$ m/s$^2$; $v_{\max} = 0.8\pi \approx 2.5$ m/s. (2) At $x = 0.20$ mm: $a = 800\pi^2 \approx 7.9 \times 10^3$ m/s$^2$; $v = 0.4\pi\sqrt{3} \approx 2.2$ m/s.
The angular frequency is $\omega = 2\pi f = 2000\pi$ rad/s and the amplitude is $A = 0.40$ mm $= 4.0 \times 10^{-4}$ m.
Part 1. $a_{\max} = \omega^2 A = (2000\pi)^2 \times 4.0 \times 10^{-4} = 1600\pi^2 \approx 1.6 \times 10^4$ m/s$^2$, and $v_{\max} = \omega A = 2000\pi \times 4.0 \times 10^{-4} = 0.8\pi \approx 2.5$ m/s.
Part 2. At displacement $x = 0.20$ mm $= 2.0 \times 10^{-4}$ m (exactly half the amplitude): the acceleration magnitude is $a = \omega^2 x = (2000\pi)^2 \times 2.0 \times 10^{-4} = 800\pi^2 \approx 7.9 \times 10^3$ m/s$^2$. The speed is $v = \omega\sqrt{A^2 - x^2} = \omega A\dfrac{\sqrt{3}}{2} = 0.8\pi \times \dfrac{\sqrt{3}}{2} = 0.4\pi\sqrt{3} \approx 2.2$ m/s.