A particle is executing `S.H.M.` of amplitude `4 cm` and `T = 4` sec. The time taken by it to move from positive extreme position to half the amplitude is

Options

A. `1 sec`
B. `1//3 sec`
C. `2//3 sec`
D. `sqrt(3//2)sec`

Explanation

To solve the problem of finding the time taken by a particle executing Simple Harmonic Motion (S.H.M.) to move from the positive extreme position to half the amplitude, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Amplitude (A) = 4 cm
- Time period (T) = 4 seconds

2. **Determine Half the Amplitude**:
- Half the amplitude (A/2) = 4 cm / 2 = 2 cm

3. **Understand the Positions in S.H.M.**:
- The positive extreme position corresponds to the maximum displacement, which is +4 cm.
- The half amplitude position corresponds to a displacement of +2 cm.

4. **Use the Equation of S.H.M.**:
- The displacement in S.H.M. can be expressed as:
\[ y = A \cos(\omega t) \]
- Here, \( y \) is the displacement, \( A \) is the amplitude, and \( \omega \) is the angular frequency.

5. **Calculate the Angular Frequency (\(\omega\))**:
- The angular frequency is given by:
\[ \omega = \frac{2\pi}{T} \]
- Substituting the value of T:
\[ \omega = \frac{2\pi}{4} = \frac{\pi}{2} \text{ rad/sec} \]

6. **Set Up the Equation for Half Amplitude**:
- We want to find the time \( t \) when the displacement \( y = 2 \) cm:
\[ 2 = 4 \cos(\omega t) \]
- Dividing both sides by 4:
\[ \frac{1}{2} = \cos(\omega t) \]

7. **Solve for \(\omega t\)**:
- The cosine function equals \( \frac{1}{2} \) at:
\[ \omega t = \frac{\pi}{3} \text{ (since cosine is positive in the first quadrant)} \]

8. **Substitute for \(\omega\)**:
- We know \( \omega = \frac{\pi}{2} \):
\[ \frac{\pi}{2} t = \frac{\pi}{3} \]

9. **Solve for \( t \)**:
- Rearranging gives:
\[ t = \frac{\frac{\pi}{3}}{\frac{\pi}{2}} = \frac{2}{3} \text{ seconds} \]

### Final Answer:
The time taken by the particle to move from the positive extreme position to half the amplitude is **\(\frac{2}{3}\) seconds**.

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