Moderate -0.8 This is a standard SHM question requiring only direct application of the formula v_max = ωa. Given ẍ = -25x, we identify ω² = 25 so ω = 5, and with amplitude a = 9m, v_max = 5 × 9 = 45 m/s. It's a single-step recall question with multiple choice format, making it easier than average despite being Further Maths content.
The movement of a particle is described by the simple harmonic equation
$$\ddot{x} = -25x$$
where \(x\) metres is the displacement of the particle at time \(t\) seconds, and \(\ddot{x}\) m s\(^{-2}\) is the acceleration of the particle.
The maximum displacement of the particle is 9 metres.
Find the maximum speed of the particle.
Circle your answer.
[1 mark]
\(15\) m s\(^{-1}\) \quad\quad \(45\) m s\(^{-1}\) \quad\quad \(75\) m s\(^{-1}\) \quad\quad \(135\) m s\(^{-1}\)
The movement of a particle is described by the simple harmonic equation
$$\ddot{x} = -25x$$
where $x$ metres is the displacement of the particle at time $t$ seconds, and $\ddot{x}$ m s$^{-2}$ is the acceleration of the particle.
The maximum displacement of the particle is 9 metres.
Find the maximum speed of the particle.
Circle your answer.
[1 mark]
$15$ m s$^{-1}$ \quad\quad $45$ m s$^{-1}$ \quad\quad $75$ m s$^{-1}$ \quad\quad $135$ m s$^{-1}$
\hfill \mbox{\textit{AQA Further Paper 2 2024 Q2 [1]}}