A particle is subjected to two simple harmonic motions.
`x_(1) = 4.0 sin (100pi t)` and `x_(2) = 3.0 sin(100pi t + (pi)/(3))`
Find
(a) the displacement at `t = 0`
(b) the maximum speed of the particle and
(c ) the maximum acceleration of the particle.

Explanation

To solve the problem step by step, we will analyze the two simple harmonic motions (SHM) given and find the required quantities.

### Given:
1. \( x_1 = 4.0 \sin(100\pi t) \)
2. \( x_2 = 3.0 \sin(100\pi t + \frac{\pi}{3}) \)

### Step 1: Find the resultant displacement at \( t = 0 \)

**Solution:**
At \( t = 0 \):
- For \( x_1 \):
\[ x_1(0) = 4.0 \sin(100\pi \cdot 0) = 4.0 \sin(0) = 0 \]

- For \( x_2 \):
\[ x_2(0) = 3.0 \sin(100\pi \cdot 0 + \frac{\pi}{3}) = 3.0 \sin\left(\frac{\pi}{3}\right) = 3.0 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \approx 2.598 \]

Now, the total displacement \( x \) at \( t = 0 \):
\[ x(0) = x_1(0) + x_2(0) = 0 + \frac{3\sqrt{3}}{2} \approx 2.598 \]

Thus, the displacement at \( t = 0 \) is approximately \( 2.6 \) units.

### Step 2: Find the maximum speed of the particle

**Solution:**
The maximum speed \( V_{\text{max}} \) in SHM is given by:
\[ V_{\text{max}} = A \omega \]
Where:
- \( A \) is the resultant amplitude
- \( \omega = 100\pi \)

To find the resultant amplitude \( A \):
Using the formula for the resultant amplitude when two SHMs are combined:
\[ A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\phi)} \]
Where:
- \( A_1 = 4 \)
- \( A_2 = 3 \)
- \( \phi = \frac{\pi}{3} \)

Calculating:
\[ A = \sqrt{4^2 + 3^2 + 2 \cdot 4 \cdot 3 \cdot \cos\left(\frac{\pi}{3}\right)} \]
\[ = \sqrt{16 + 9 + 2 \cdot 4 \cdot 3 \cdot \frac{1}{2}} \]
\[ = \sqrt{16 + 9 + 12} = \sqrt{37} \approx 6.08 \]

Now substituting \( A \) and \( \omega \):
\[ V_{\text{max}} = 6.08 \cdot 100\pi \approx 1917.88 \text{ units} \]

### Step 3: Find the maximum acceleration of the particle

**Solution:**
The maximum acceleration \( A_{\text{max}} \) in SHM is given by:
\[ A_{\text{max}} = A \omega^2 \]
Substituting the values:
\[ A_{\text{max}} = 6.08 \cdot (100\pi)^2 \]
Calculating \( (100\pi)^2 \):
\[ (100\pi)^2 = 10000\pi^2 \approx 98696.44 \]
Thus,
\[ A_{\text{max}} \approx 6.08 \cdot 98696.44 \approx 600000 \text{ units} \]

### Final Answers:
(a) The displacement at \( t = 0 \) is approximately \( 2.6 \) units.
(b) The maximum speed of the particle is approximately \( 1917.88 \) units.
(c) The maximum acceleration of the particle is approximately \( 600000 \) units.