An iron sphere of mass 40 g hangs from a spring and oscillates with simple harmonic motion - Leaving Cert Physics - Question a - 2021

To calculate the spring constant (k), we start with the relationship between the period (T) and the spring constant:

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

Rearranging, we have:

$k = \frac{4\pi^2 m}{T^2}$

Given:

• Mass (m) = 40 g = 0.04 kg (conversion from grams to kilograms)
• Period (T) = 0.74 s

Calculating:

$k = \frac{4\pi^2 \times 0.04}{(0.74)^2} \approx 2.88 \text{ N m}^{-1}$

Using the formula for acceleration in simple harmonic motion:

$a = \omega^2 x$

We first need to calculate the angular frequency ($\omega$):

$\omega = \frac{2\pi}{T}$

Calculating:

$\omega = \frac{2\pi}{0.74} \approx 8.49 \text{ s}^{-1}$

Now we can find the acceleration:

• Displacement (x) = 18 mm = 0.018 m

Substituting into the acceleration formula:

$a = (8.49)^2 \times 0.018 \approx 1.3 \text{ m s}^{-2}$

When the magnet is attached, the total force acting on the system can be given as:

$mg = kx$

Where:

• m = mass of the magnet
• k = spring constant (from previous calculation, k = 2.88 N m⁻¹)
• x = extension of the spring = 15 mm = 0.015 m

Rearranging the equation gives:

$m = \frac{kx}{g}$

Substituting values:

$m = \frac{2.88 \times 0.015}{9.8} \approx 0.0044 \text{ kg} \approx 4.4 ext{ g}$