Step 1: Understanding the Concept:
In Simple Harmonic Motion (SHM), the displacement of a particle varies as a sine or cosine function of time.
The particle moves fastest at the mean position and comes to a momentary halt at the extreme positions.
Because the velocity is not constant, the time taken to cover equal distances is not the same throughout the path.
It will take less time to cover the first half of the amplitude (near the mean position) than the second half (near the extreme).
Step 2: Key Formula or Approach:
The general equation for displacement starting from the mean position (\(x=0\)) at \(t=0\) is:
\[ x(t) = A \sin(\omega t) \]
Where:
\(A\) is the amplitude.
\(\omega\) is the angular frequency, defined as \(\omega = \frac{2\pi}{T}\).
Step 3: Detailed Explanation:
The problem asks for the time \(t\) when the displacement \(x\) reaches \(\frac{A}{2}\).
Substitute the value of \(x\) into the displacement equation:
\[ \frac{A}{2} = A \sin(\omega t) \]
Divide both sides by \(A\):
\[ \frac{1}{2} = \sin(\omega t) \]
We know from trigonometry that \(\sin(\theta) = \frac{1}{2}\) when \(\theta = \frac{\pi}{6}\) radians (or $30^{\circ}$).
Therefore:
\[ \omega t = \frac{\pi}{6} \]
Now, substitute \(\omega = \frac{2\pi}{T}\):
\[ \left( \frac{2\pi}{T} \right) t = \frac{\pi}{6} \]
Divide both sides by \(\pi\):
\[ \frac{2t}{T} = \frac{1}{6} \]
Solve for \(t\):
\[ t = \frac{T}{2 \times 6} \]
\[ t = \frac{T}{12} \]
Comparison:
- Time to go from \(0\) to \(A/2\) is \(T/12\).
- Time to go from \(0\) to \(A\) is \(T/4\).
- Thus, time to go from \(A/2\) to \(A\) is \(T/4 - T/12 = 2T/12 = T/6\).
As expected, it takes twice as long to cover the outer half of the amplitude because the particle is slowing down.
Step 4: Final Answer:
The time taken is \(T/12\).