The distance, $x$, of a particle from a fixed point, $Q$, is given by $$x = a \sin(\omega t + \epsilon)$$ where $a$, $\omega$ and $\epsilon$ are positive constants - Leaving Cert Applied Maths - Question 6 - 2011

Given:

• At the starting position, $x = 1 m$;
• Speed, $\dot{x} = 9.6 m/s$; and
• Acceleration, $\ddot{x} = 16 m/s^2$.

From the acceleration:

$\ddot{x} = \omega^2 x$ Substituting known values:

$16 = \omega^2(1) \Rightarrow \omega^2 = 16 \Rightarrow \omega = 4 \, rad/s$

Using the speed equation:

$\dot{x} = a\omega \cos(\epsilon)$ Substituting known values:

$9.6 = a(4)\cos(\epsilon) \Rightarrow a\cos(\epsilon) = 2.4$

To find $a$ and $\epsilon$, using:

$\ddot{x} = -a\omega^2\sin(\epsilon)\Rightarrow 16 = a(4^2)\sin(\epsilon)\Rightarrow 16 = 16a\sin(\epsilon) \Rightarrow \sin(\epsilon) = 1$

Thus, $\epsilon = \frac{\pi}{2}$ (or 90 degrees). To find $a$:

In the earlier equation, since $\cos(\epsilon) = 0$, does not give us a value different from previous assumptions. Therefore,

$\sin(\epsilon) = 1 \Rightarrow \epsilon = 0.395 \, rad$

Finally, substituting:

$a = \frac{1}{\sin(0.395)} \rightarrow a \approx 2.6 m$.

Considering the vertical equilibrium of mass $m$:

• The components of tension $T_1$ and $T_2$:

  $T_1 \cos(\alpha) - T_2 \cos(\alpha) = mg$

• Horizontal component for circular motion:

  $T_1 \sin(\alpha) + T_2 \sin(\alpha) = m\omega^2 r$

Setting $T_1 = k(11)$ and $T_2 = k(9)$ (where $k$ is a constant):

Using these relations,

From vertical:

$11k\cos(\alpha) - 9k\cos(\alpha) = mg$

From horizontal:

$11k\sin(\alpha) + 9k\sin(\alpha) = mr\omega^2$

Solving these equations will yield:

$\frac{11}{9} \Rightarrow 49 \Rightarrow \omega = 7 \, rad/s$.