Compare with the standard SHM equation

For simple harmonic motion, $y = A\sin(\omega t + \phi)$ where $\omega$ is the angular frequency.

Given: $y = A\sin(\pi t + \phi)$ so, $\omega = \pi\ \text{rad s}^{-1}$

Use the relation between angular frequency and time period

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

Therefore, $\pi = \frac{2\pi}{T}$ $T = 2\ \text{s}$

Apply the time period formula for a simple pendulum

For a simple pendulum, $T = 2\pi\sqrt{\frac{l}{g}}$

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

Divide by 2: $1 = \pi\sqrt{\frac{l}{g}}$

$\sqrt{\frac{l}{g}} = \frac{1}{\pi}$

Squaring both sides: $\frac{l}{g} = \frac{1}{\pi^2}$

Hence, $l = \frac{g}{\pi^2}$

Substitute $g = 9.8\ \text{m s}^{-2}$

$l = \frac{9.8}{\pi^2}$

Using $\pi^2 \approx 9.8696$, $l \approx \frac{9.8}{9.8696} \approx 0.993\ \text{m}$

Converting to cm: $0.993\ \text{m} = 99.3\ \text{cm}$

This matches closest with: $\boxed{99.4\ \text{cm}}$

Check options

• A: $97.23$ cm — incorrect
• B: $25.3$ cm — incorrect
• C: $99.4$ cm — correct
• D: $406.1$ cm — incorrect