A particle is subjected to two simple harmonic motions given by
`x_(1) = 2.0sin (100 pi t)` and `x_(2) = 2.0sin (120pi t + pi //3)`
where, `x` is in `cm` and `t` in second. Find the displacement of the particle at
(a) `t = 0.0125`,
(b) `t = 0.025`.

Explanation

To find the displacement of a particle subjected to two simple harmonic motions at given times, we can follow these steps:

### Given:
1. The two simple harmonic motions are:
- \( x_1 = 2.0 \sin(100 \pi t) \)
- \( x_2 = 2.0 \sin(120 \pi t + \frac{\pi}{3}) \)

### Step 1: Calculate Displacement at \( t = 0.0125 \) seconds

#### a. Calculate \( x_1 \) at \( t = 0.0125 \):
\[ x_1 = 2.0 \sin(100 \pi \times 0.0125) \]
\[ = 2.0 \sin(1.25 \pi) \]
\[ = 2.0 \times (-1) = -2.0 \text{ cm} \]

#### b. Calculate \( x_2 \) at \( t = 0.0125 \):
\[ x_2 = 2.0 \sin(120 \pi \times 0.0125 + \frac{\pi}{3}) \]
\[ = 2.0 \sin(1.5 \pi + \frac{\pi}{3}) \]
\[ = 2.0 \sin(1.5 \pi + 1.0472) \quad (\text{since } \frac{\pi}{3} \approx 1.0472) \]
\[ = 2.0 \sin(1.5 \pi + 1.0472) = 2.0 \sin(2.6172) \approx 2.0 \times (-0.5) = -1.0 \text{ cm} \]

#### c. Calculate Total Displacement \( x \):
\[ x = x_1 + x_2 = -2.0 \text{ cm} + (-1.0 \text{ cm}) = -3.0 \text{ cm} \]

### Step 2: Calculate Displacement at \( t = 0.025 \) seconds

#### a. Calculate \( x_1 \) at \( t = 0.025 \):
\[ x_1 = 2.0 \sin(100 \pi \times 0.025) \]
\[ = 2.0 \sin(2.5 \pi) \]
\[ = 2.0 \times 1 = 2.0 \text{ cm} \]

#### b. Calculate \( x_2 \) at \( t = 0.025 \):
\[ x_2 = 2.0 \sin(120 \pi \times 0.025 + \frac{\pi}{3}) \]
\[ = 2.0 \sin(3.0 \pi + \frac{\pi}{3}) \]
\[ = 2.0 \sin(3.0 \pi + 1.0472) = 2.0 \sin(3.0 \pi + 1.0472) = 2.0 \sin(4.0472) \approx 2.0 \times (-0.5) = -1.73 \text{ cm} \]

#### c. Calculate Total Displacement \( x \):
\[ x = x_1 + x_2 = 2.0 \text{ cm} + (-1.73 \text{ cm}) = 0.27 \text{ cm} \]

### Final Answers:
- (a) At \( t = 0.0125 \) seconds, the displacement is \( -3.0 \) cm.
- (b) At \( t = 0.025 \) seconds, the displacement is \( 0.27 \) cm.