A spring of natural length 150 mm obeys Hooke's law - Leaving Cert Physics - Question 7 - 2022

To calculate the elastic constant, we first find the extension of the spring.

Initial length = 150 mm Final length under load = 185 mm Extension, $x = 185 ext{ mm} - 150 ext{ mm} = 35 ext{ mm} = 0.035 ext{ m}$

The weight of the object is: $W = mg = (0.2 ext{ kg})(9.81 ext{ m/s}^2) = 1.962 ext{ N}$

Now applying Hooke's law: $F = kx \implies k = \frac{F}{x} = \frac{1.962}{0.035} \approx 56 ext{ N/m}$

The period of oscillation for a mass-spring system is given by the formula:

$T = 2\pi \sqrt{\frac{m}{k}}$

Where:

• $m = 0.2 ext{ kg}$
• $k = 56 ext{ N/m}$

So, $T = 2\pi \sqrt{\frac{0.2}{56}} \approx 0.375 ext{ s}$

The maximum acceleration in simple harmonic motion is given by:

$a_{max} = \omega^2 A$

Where:

• $A = 0.2 ext{ m} - 0.15 ext{ m} = 0.05 ext{ m}$
• Angular frequency, $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.375}$

Calculating: $\omega \approx 16.73 ext{ rad/s}$

Thus, $a_{max} = (16.73)^2 (0.05) \approx 14.03 ext{ m/s}^2$

At maximum acceleration, the speed of the object is zero because it is at the extreme positions of its oscillation. Therefore, the speed when maximum acceleration occurs is:

$v = 0 ext{ m/s}$