A string of total length 30 cm and linear mass density 0.5 g cm^ -1...

For a string of length $L$ secured to a fixed wall at one end and a frictionless ring (free end) at the other, the boundary conditions dictate how a wave pulse reflects:
1. At the free end (ring), the pulse reflects without any phase change (an upright pulse remains upright).
2. At the fixed end (wall), the pulse reflects with a phase change of $\pi$ (an upright pulse becomes inverted).

Let us trace the journey of the initial upright pulse to determine when it returns to its exact initial state (same position, same shape, and same direction of motion):
- The pulse travels towards the free end and reflects. It remains upright but reverses direction.
- It then travels to the fixed end and reflects. It becomes inverted and reverses direction.
- It travels back to the free end and reflects. It remains inverted.
- It travels to the fixed end and reflects again. It becomes upright and reverses direction, now moving in its original direction.

To complete one full cycle and return to its exact initial state, the pulse must undergo two reflections at the free end and two reflections at the fixed end. This corresponds to traveling a total distance of $4L$.

Given:
Length of the string, $L = 30 \text{ cm}$
Wave speed, $v = 20 \text{ cm s}^{-1}$

The time period $T$ of the periodic change in the shape of the string is the time taken to travel this total distance of $4L$:

$T = \dfrac{4L}{v}$

Substituting the given values:

$T = \dfrac{4 \times 30 \text{ cm}}{20 \text{ cm s}^{-1}} = \dfrac{120}{20} \text{ s} = 6 \text{ s}$

Answer: $6 \text{ s}$