A simple pendulum can execute simple harmonic motion - Leaving Cert Physics - Question a - 2018

A simple pendulum executes simple harmonic motion when it swings at small angles (typically less than 15 degrees). At larger angles, the motion deviates from simple harmonic behavior due to increased non-linearity in the restoring force.

The period (T) of a simple pendulum is directly related to its length (L) by the formula: $T = 2\pi\sqrt{\frac{L}{g}}$ where g is the acceleration due to gravity. This indicates that the period is proportional to the square root of the length of the pendulum.

Given the mass (m) = 60 g = 0.06 kg and the period (T) = 0.85 s, we can first find the angular frequency ($\omega$):

$T = \frac{2\pi}{\omega} \implies \omega = \frac{2\pi}{T} \approx 7.39 \text{ rad/s}$

Then we use the formula to relate angular frequency to the spring constant:

$\omega^2 = \frac{k}{m} \implies k = m\omega^2 = 0.06 \times (7.39)^2 \approx 3.28 \, \text{N m}^{-1}$

Using Hooke's Law:

$F = -k s$

Where s is the extension of the spring. Since the mass is at rest, the force exerted by the mass (F = mg) is equal to the spring force:

$F = mg = 0.06 kg \times 9.8 m/s^2 \approx 0.588 \, \text{N}$

Setting this equal to the spring force:

$0.588 = k s \implies s = \frac{0.588}{3.28} \approx 0.18 \, \text{m}$

The natural length of the spring is 0.50 m, thus the length when the mass is at rest is:

$l = \text{natural length} + s = 0.50 + 0.18 = 0.68 \, \text{m}$