A particle of mass 0.5 kg is suspended from a fixed point P by a spring which executes simple harmonic motion with amplitude 0.2 m - Leaving Cert Applied Maths - Question 6 - 2012

Using Newton's second law, the force exerted by the spring can be computed as:

The weight of the particle is given by: $mg = 0.5 imes 9.8 = 4.9 ext{ N}$

The net force equation is: $T - mg = ma$ Substituting for ( a ): $T - 4.9 = 0.5 \times \frac{\pi^2}{5}$ Solving for ( T ): $T = 4.9 + 0.5 \times \frac{\pi^2}{5}$ Calculating this: $T \approx 4.9 + 0.5 \times 1.97 \approx 4.9 + 0.985 = 5.885 \approx 5.9 ext{ N (to one place of decimals)}$

Using conservation of energy between points A and B for the sphere:

At point A, potential energy is given by: $PE = mgh = mg(0.3 - 0.3\cos\alpha)$ At point B, kinetic energy is: $KE = \frac{1}{2}mv^2$ Setting them equal (using ( g = 10 \text{ m/s}^2 ) for simplicity): $\frac{1}{2}mv^2 = mg(0.3 - 0.3\cos\alpha)$ Canceling ( m ) and substituting values yields: $v^2 = 0.6g(1 - \cos\alpha)$ Using (\cos \alpha = 2(1 - \cos \alpha)): $$v^2 = 0.3g = 0.3 \times 10 = 3 ext{ hence } v = \sqrt{3} ext{ m/s}$.

The horizontal distance after time ( t ) is expressed as:

$x = 0.3\sin\alpha + 1.4\cos\alpha\times t$ We substitute the known values: $\sqrt{\frac{5}{10}} + kt = \frac{\sqrt{5}}{10} + kt$ Therefore, equating gives us: $k = \frac{14}{15}$ Thus, ( k \approx 0.93 \text{ (to two decimal places)}$$