(a) The distance, $x$, of a particle from a fixed point, $o$, is given by $$x = a \, \cos(\omega t + \epsilon)$$ where $a$, $\omega$ and $\epsilon$ are constants - Leaving Cert Applied Maths - Question 6 - 2009

Given the period $T = 11$ seconds, we can find the angular frequency:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{11} \approx \frac{4}{7} \text{ rad/s}$

The maximum speed $v_{max}$ in simple harmonic motion is given by:

$v_{max} = \omega a$

Next, we need to find the amplitude $a$. We know:

• The particle's starting position is 5 cm from the mean position.
• Using the cosine function:

$\cos(\omega t + \epsilon) = \frac{x}{a}$

Substituting $x = 5$ cm yields:

$5 = a \cos(0) \Rightarrow a = 5 \text{ cm}$

Thus, we find:

$v_{max} = \frac{4}{7} \cdot 5 \approx 2.86 \text{ cm/s}$

To round this to two decimal places, we find:

$v_{max} \approx 2.86 ext{ cm/s}$

To find the greatest allowable amplitude, we start by determining the frequency of the oscillations:

$f = \frac{N}{60} ext{ Hz}$

The angular frequency is:

$\omega = 2\pi f = \frac{2\pi N}{60} = \frac{\pi N}{30}$

The force on the object on the table is given by:

$F = ma = m \omega^2 r$

Where $r$ represents the amplitude of motion. For the object not to slip, the frictional force must meet the following condition:

$\mu R = mg$

Thus, setting up the equation based on the maximum friction force:

$\mu R = mr\omega^2$

Combining these gives:

$R = \mu mg\cdot \frac{1}{r \omega^2}$

Substituting $\omega$ into the equation yields:

$\mu R = m r \left(\frac{\pi N}{30}\right)^2$

Rearranging gives:

$r = \frac{900g\mu}{\pi^2 N^2}$

Thus, the greatest allowable amplitude in terms of $\mu$ and $N$ is:

$r = \frac{900g\mu}{\pi^2 N^2}$