A mass of 400 g is attached to a vertical spiral spring and set into vibration. It it's period is 2.0 s, calculate :

The given mass is $m = 400 \text{ g}$.

$$m = 400 \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 0.400 \text{ kg}$$

The period $T$ of a mass-spring system is given by the formula:

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

where $m$ is the mass and $k$ is the spring constant.

Square both sides of the equation:

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

Now, solve for $k$:

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

Given $m = 0.400 \text{ kg}$ and $T = 2.0 \text{ s}$.

$$k = \frac{4\pi^2 (0.400 \text{ kg})}{(2.0 \text{ s})^2}$$ $$k = \frac{4\pi^2 (0.400)}{4.0}$$ $$k = \pi^2 (0.400)$$

Using $\pi \approx 3.14159$:

$$k \approx (3.14159)^2 \times 0.400$$ $$k \approx 9.8696 \times 0.400$$ $$k \approx 3.94784 \text{ N/m}$$

Rounding to two significant figures (based on the period):

$$k \approx 3.9 \text{ N/m}$$

The spring constant is:

$$\text{3.9 N/m}$$