Step 1: Understanding the Concept:
For a particle executing Simple Harmonic Motion (SHM) starting from the mean position, its displacement is given by a sine function.
We need to calculate its position at $t=1\text{ s}$ and $t=2\text{ s}$ to determine the distances covered in the specific time intervals.
Step 2: Key Formula or Approach:
Equation of motion: $y(t) = A \sin(\omega t)$
Angular frequency: $\omega = \frac{2\pi}{T}$
Distance in 1st second ($d_1$): $|y(1) - y(0)|$
Distance in 2nd second ($d_2$): $|y(2) - y(1)|$ (assuming it hasn't reversed direction, which it doesn't since $t=2\text{s} = T/4$).
Step 3: Detailed Explanation:
Given values:
Time period, $T = 8\text{ s}$
Amplitude, $A = 4\sqrt{2}\text{ m}$
Calculate angular frequency: $\omega = \frac{2\pi}{8} = \frac{\pi}{4}\text{ rad/s}$.
The equation for displacement is: $y(t) = 4\sqrt{2} \sin\left(\frac{\pi}{4} t\right)$.
- At $t = 0\text{ s}$, $y(0) = 0$.
- At $t = 1\text{ s}$ (end of 1st second):
$y(1) = 4\sqrt{2} \sin\left(\frac{\pi}{4} \times 1\right) = 4\sqrt{2} \sin(45^\circ)$
$y(1) = 4\sqrt{2} \times \left(\frac{1}{\sqrt{2}}\right) = 4\text{ m}$.
- At $t = 2\text{ s}$ (end of 2nd second):
$y(2) = 4\sqrt{2} \sin\left(\frac{\pi}{4} \times 2\right) = 4\sqrt{2} \sin\left(\frac{\pi}{2}\right)$
$y(2) = 4\sqrt{2} \times 1 = 4\sqrt{2}\text{ m}$.
Now, calculate the distances:
Distance travelled in $1^{\text{st}}$ second ($d_1$) is the change in position from $t=0$ to $t=1$:
\[ d_1 = y(1) - y(0) = 4 - 0 = 4\text{ m} \]
Distance travelled in $2^{\text{nd}}$ second ($d_2$) is the change in position from $t=1$ to $t=2$:
\[ d_2 = y(2) - y(1) = 4\sqrt{2} - 4 = 4(\sqrt{2} - 1)\text{ m} \]
Find the ratio $d_1 : d_2$:
\[ \text{Ratio} = \frac{4}{4(\sqrt{2} - 1)} = \frac{1}{\sqrt{2} - 1} \]
Step 4: Final Answer:
The ratio is $1 : (\sqrt{2} - 1)$.