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))`
(c) `x = A` to `x = A//2`
(d) `x = -(A)/(sqrt(2))` "to" `x = (A)/(sqrt(2))`
(e) `x = (A)/(sqrt(2))` to `x = A`.

Answer

To solve the problem of finding the time required for a simple harmonic motion (SHM) to travel between specified positions, we will use the relationship between displacement, amplitude, and time period.

### Step-by-Step Solution:

1. **Understanding SHM**:
- In simple harmonic motion, the displacement \( x \) from the mean position can be expressed as:
\[ x = A \sin(\omega t) \]
where \( \omega = \frac{2\pi}{T} \) is the angular frequency, and \( A \) is the amplitude.

2. **Finding Time for Each Case**:
- We will calculate the time taken to travel from one position to another based on the angle subtended at the center of the circular motion.

#### (a) From \( x = 0 \) to \( x = \frac{A}{2} \):
- At \( x = 0 \), \( \sin(\theta) = 0 \) (which corresponds to \( \theta = 0 \)).
- At \( x = \frac{A}{2} \), \( \sin(\theta) = \frac{1}{2} \) (which corresponds to \( \theta = 30^\circ \)).
- The time taken to move from \( 0 \) to \( 30^\circ \) is:
\[ t = \frac{T}{360^\circ} \times 30^\circ = \frac{T}{12} \]

#### (b) From \( x = 0 \) to \( x = \frac{A}{\sqrt{2}} \):
- At \( x = 0 \), \( \theta = 0 \).
- At \( x = \frac{A}{\sqrt{2}} \), \( \sin(\theta) = \frac{1}{\sqrt{2}} \) (which corresponds to \( \theta = 45^\circ \)).
- The time taken is:
\[ t = \frac{T}{360^\circ} \times 45^\circ = \frac{T}{8} \]

#### (c) From \( x = A \) to \( x = \frac{A}{2} \):
- At \( x = A \), \( \theta = 90^\circ \).
- At \( x = \frac{A}{2} \), \( \theta = 30^\circ \).
- The angle subtended is \( 90^\circ - 30^\circ = 60^\circ \).
- The time taken is:
\[ t = \frac{T}{360^\circ} \times 60^\circ = \frac{T}{6} \]

#### (d) From \( x = -\frac{A}{\sqrt{2}} \) to \( x = \frac{A}{\sqrt{2}} \):
- At \( x = -\frac{A}{\sqrt{2}} \), \( \theta = -45^\circ \) (or \( 315^\circ \)).
- At \( x = \frac{A}{\sqrt{2}} \), \( \theta = 45^\circ \).
- The angle subtended is \( 45^\circ - (-45^\circ) = 90^\circ \).
- The time taken is:
\[ t = \frac{T}{360^\circ} \times 90^\circ = \frac{T}{4} \]

#### (e) From \( x = \frac{A}{\sqrt{2}} \) to \( x = A \):
- At \( x = \frac{A}{\sqrt{2}} \), \( \theta = 45^\circ \).
- At \( x = A \), \( \theta = 90^\circ \).
- The angle subtended is \( 90^\circ - 45^\circ = 45^\circ \).
- The time taken is:
\[ t = \frac{T}{360^\circ} \times 45^\circ = \frac{T}{8} \]

### Summary of Results:
- (a) \( \frac{T}{12} \)
- (b) \( \frac{T}{8} \)
- (c) \( \frac{T}{6} \)
- (d) \( \frac{T}{4} \)
- (e) \( \frac{T}{8} \)