A simple harmonic motion has an amplitude `A` and time period `T`. Find the time required bu it to trvel directly form
(a) `x = 0` to `x = A//2`
(b) `x = 0` to `x = (A)/(sqrt(2))`

Answer

To solve the problem, we will analyze the motion of a simple harmonic oscillator (SHO) and derive the time taken to travel from specific positions.

### Given:
- Amplitude \( A \)
- Time period \( T \)

### (a) Time required to travel from \( x = 0 \) to \( x = \frac{A}{2} \)

1. **Displacement Equation**:
The displacement in simple harmonic motion can be expressed as:
\[ x(t) = A \cos(\omega t) \]
where \( \omega = \frac{2\pi}{T} \).

2. **Set up the equation**:
For \( x = \frac{A}{2} \):
\[ \frac{A}{2} = A \cos(\omega t) \]
Dividing both sides by \( A \):
\[ \frac{1}{2} = \cos(\omega t) \]

3. **Find \( \omega t \)**:
Taking the inverse cosine:
\[ \omega t = \cos^{-1}\left(\frac{1}{2}\right) \]
We know that:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
Therefore:
\[ \omega t = \frac{\pi}{3} \]

4. **Substituting \( \omega \)**:
Substitute \( \omega = \frac{2\pi}{T} \):
\[ \frac{2\pi}{T} t = \frac{\pi}{3} \]

5. **Solve for \( t \)**:
Rearranging gives:
\[ t = \frac{T}{6} \]

### (b) Time required to travel from \( x = 0 \) to \( x = \frac{A}{\sqrt{2}} \)

1. **Set up the equation**:
For \( x = \frac{A}{\sqrt{2}} \):
\[ \frac{A}{\sqrt{2}} = A \cos(\omega t) \]
Dividing both sides by \( A \):
\[ \frac{1}{\sqrt{2}} = \cos(\omega t) \]

2. **Find \( \omega t \)**:
Taking the inverse cosine:
\[ \omega t = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \]
We know that:
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Therefore:
\[ \omega t = \frac{\pi}{4} \]

3. **Substituting \( \omega \)**:
Substitute \( \omega = \frac{2\pi}{T} \):
\[ \frac{2\pi}{T} t = \frac{\pi}{4} \]

4. **Solve for \( t \)**:
Rearranging gives:
\[ t = \frac{T}{8} \]

### Final Answers:
- (a) The time required to travel from \( x = 0 \) to \( x = \frac{A}{2} \) is \( \frac{T}{6} \).
- (b) The time required to travel from \( x = 0 \) to \( x = \frac{A}{\sqrt{2}} \) is \( \frac{T}{8} \).