Question 14.2: i) Show that x = 10 sin 3t satisfies the simple harmonic mot...
i) Show that x = 10 sin 3t satisfies the simple harmonic motion equation
\ddot{x} = -9x.
ii) Sketch the graph of x = 10 sin 3t and deduce the amplitude, period and frequency of this motion.
iii) Verify that \dot{x}^{2} = v² = ω²(a² – x²)
iv) Show that x = 25 sin 3t also satisfies \dot{x} = -9x and comment on this result.
i) x = 10 sin 3t
Differentiating with respect to t to find the velocity and acceleration:
v = \dot{x} = 30 cos 3t ①
and \ddot{x} = -90 sin 3t
⇒ \dot{x} = -9x ②
This is the SHM equation given in the question.
ii) The graph of the function is shown below.
The graph shows that the value of x always lies between -10 and 10, so the amplitude of the motion is 10 units.
Each cycle repeats in the time \frac{2π}{3} so the period of the oscillations, denoted by T, is \frac{2π}{3} . The frequency, f, is \frac{1}{T} or \frac{3}{2π}.
iii) From ① \dot{x}^{2} = 900 cos^{2} 3t
\begin{matrix} = 900(1 – sin² 3t) &\boxed{ω² = 9 \text{ from equation ②}} \end{matrix}
= 9(100 – 100 sin² 3t)
= ω²(a² – x²).
iv) Differentiating the function x = 25 sin 3t twice gives:
\dot{x} = 75 cos 3t
\ddot{x} = -225 sin 3t
= -9 (25 sin 3t)
⇒ \ddot{x} = -9x as required
This shows that the amplitude of the oscillation is not determined by the differential equation. In this case it can be 10 or 25, or indeed have any other value.
