Funda lendatshana elandelayo bese uphendula imibuzo elandelayo:

1.1 Sihinu ifọ ti akọkọ ti ìsalẹ̀ (State the formula for the period of oscillation):

Step 1.1: The period $T$ of a simple pendulum undergoing small oscillations is given by

$T = 2\pi \sqrt{\frac{l}{g}}$

where $l$ is the length of the string in meters (m) and $g$ is the acceleration due to gravity in meters per second squared ($m/s^2$).

T = 2\pi \sqrt{\dfrac{l{g}}}

1.2 Nika ìyè g ní South Africa (Value of $g$ in South Africa):

Step 1.2: The acceleration due to gravity in South Africa is approximately

$g \approx 9.8 \, m/s^2$

9.8 , \mathrm{m/s^2}

1.3 Ṣe ìṣirò ìyè akọkọ fún pendulum tí ó ni gigùn 1 m (Calculate period for length 1 m):

Step 1.3: Use the formula for the period:

$T = 2\pi \sqrt{\frac{l}{g}}$

Substitute $l = 1 \, m$ and $g = 9.8 \, m/s^2$:

$T = 2\pi \sqrt{\frac{1}{9.8}}$

Calculate the fraction:

$\frac{1}{9.8} \approx 0.1020$

Take the square root:

$\sqrt{0.1020} \approx 0.3194$

Multiply by $2\pi$ (using $\pi \approx 3.142$):

$2 \times 3.142 \times 0.3194 \approx 6.284 \times 0.3194 \approx 2.01 \, s \approx 2.0 \, s$

2.0 , \mathrm{s}

1.4 Ìṣirò fún gigùn mìíràn (Assuming length 0.5 m as common follow-up; adjust if diagram specifies):

Step 1.4: Use the formula

$T = 2\pi \sqrt{\frac{l}{g}}$

Substitute $l = 0.5 \, m$, $g = 9.8 \, m/s^2$:

$T = 2\pi \sqrt{\frac{0.5}{9.8}}$

Calculate the fraction:

$\frac{0.5}{9.8} \approx 0.05102$

Take the square root:

$\sqrt{0.05102} \approx 0.2259$

Multiply by $2\pi$:

$2 \times 3.142 \times 0.2259 \approx 6.284 \times 0.2259 \approx 1.42 \, s$

1.4 , \mathrm{s}