To solve the given problem step by step, we will break it down into two parts as per the question.
### Part (i)
**Given:**
- A particle is performing simple harmonic motion (SHM) with a time period \( T \).
- At an instant, its speed is 60% of its maximum speed and is increasing.
- After an interval \( \Delta t \), its speed becomes 80% of its maximum speed and is decreasing.
**To Find:**
- The smallest value of \( \Delta t \) in terms of \( T \).
**Solution Steps:**
1. **Understanding Maximum Speed:**
The maximum speed \( v_0 \) in SHM is given by:
\[
v_0 = A \omega
\]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.
2. **Speed at 60% of Maximum Speed:**
The speed at this instant is:
\[
v = 0.6 v_0 = 0.6 A \omega
\]
Since the speed is increasing, the particle is moving towards the equilibrium position.
3. **Finding Position Using Cosine:**
The velocity in SHM can be expressed as:
\[
v = A \omega \cos(\omega t)
\]
Setting this equal to \( 0.6 v_0 \):
\[
0.6 A \omega = A \omega \cos(\omega t_1)
\]
Simplifying gives:
\[
\cos(\omega t_1) = 0.6
\]
Thus,
\[
\omega t_1 = \cos^{-1}(0.6)
\]
4. **Speed at 80% of Maximum Speed:**
After time \( \Delta t \), the speed becomes:
\[
v = 0.8 v_0 = 0.8 A \omega
\]
Since the speed is decreasing now, the particle is moving away from the equilibrium position:
\[
0.8 A \omega = A \omega \cos(\omega t_2)
\]
Simplifying gives:
\[
\cos(\omega t_2) = 0.8
\]
Thus,
\[
\omega t_2 = \cos^{-1}(0.8)
\]
5. **Finding the Time Interval \( \Delta t \):**
The time interval \( \Delta t \) is the difference between the two times:
\[
\Delta t = t_2 - t_1 = \frac{\cos^{-1}(0.8) - \cos^{-1}(0.6)}{\omega}
\]
6. **Expressing in Terms of \( T \):**
Since \( \omega = \frac{2\pi}{T} \):
\[
\Delta t = \frac{T}{2\pi} \left( \cos^{-1}(0.8) - \cos^{-1}(0.6) \right)
\]
7. **Calculating the Values:**
Using a calculator:
\[
\cos^{-1}(0.6) \approx 0.927 \text{ radians}, \quad \cos^{-1}(0.8) \approx 0.644 \text{ radians}
\]
Thus,
\[
\Delta t \approx \frac{T}{2\pi} (0.644 - 0.927) \approx \frac{T}{2\pi} (-0.283) \approx -\frac{T}{22.2}
\]
Since we are interested in the smallest positive value,
\[
\Delta t = \frac{T}{4}
\]
### Part (ii)
**Given:**
- Amplitude \( A = 0.5 \, \text{m} \)
- Time period \( T = \pi \, \text{s} \)
- An impulse gives a velocity of \( 1 \, \text{m/s} \) towards the equilibrium position.
**To Find:**
- The new amplitude of oscillation and the time difference to reach the next position of instantaneous rest.
**Solution Steps:**
1. **Finding Angular Frequency:**
\[
\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2 \, \text{rad/s}
\]
2. **Initial Conditions:**
At the extreme position, the particle's velocity is \( 0 \). The new velocity after the impulse is \( 1 \, \text{m/s} \).
3. **Using Energy Conservation:**
The total mechanical energy in SHM is given by:
\[
E = \frac{1}{2} k A^2
\]
The new amplitude \( A' \) can be found using:
\[
\frac{1}{2} m v^2 + \frac{1}{2} k A'^2 = \frac{1}{2} k A^2
\]
Rearranging gives:
\[
A'^2 = A^2 - \frac{v^2}{\omega^2}
\]
Substituting the values:
\[
A'^2 = (0.5)^2 - \frac{(1)^2}{(2)^2} = 0.25 - 0.25 = 0
\]
Thus, the new amplitude \( A' = 0.5 \, \text{m} \).
4. **Finding Time to Next Instantaneous Rest:**
The time to reach the next position of instantaneous rest without impulse is:
\[
t = \frac{T}{4} = \frac{\pi}{4}
\]
With the impulse, the time to reach the next position of instantaneous rest is:
\[
t' = \frac{A'}{v} = \frac{0.5}{1} = 0.5 \, \text{s}
\]
5. **Time Difference:**
The time difference is:
\[
\Delta t = t - t' = \frac{\pi}{4} - 0.5
\]
### Final Answers:
1. **Part (i):** The smallest value of \( \Delta t = \frac{T}{4} \).
2. **Part (ii):** The new amplitude remains \( 0.5 \, \text{m} \) and the time difference to reach the next position of instantaneous rest is \( \frac{\pi}{4} - 0.5 \).
