A particle is subjected to two simple harmonic motions given by
`x_1=2.0sin(100pit)and x_2=2.0sin(120pit+pi/3)`, where x is in centimeter and t in second. Find the displacement of the particle at a. t=0.0125, b. t= 0.025.

Explanation

To solve the problem, we need to find the total displacement of a particle subjected to two simple harmonic motions at two different times. The equations of motion are given as:

1. \( x_1 = 2.0 \sin(100 \pi t) \)
2. \( x_2 = 2.0 \sin(120 \pi t + \frac{\pi}{3}) \)

We will calculate the displacement at \( t = 0.0125 \) seconds and \( t = 0.025 \) seconds.

### Step-by-Step Solution

**Step 1: Calculate displacement at \( t = 0.0125 \) seconds.**

1. Substitute \( t = 0.0125 \) into \( x_1 \):
\[ x_1 = 2.0 \sin(100 \pi \times 0.0125) = 2.0 \sin(1.25 \pi) \]
Since \( \sin(1.25 \pi) = -\frac{\sqrt{2}}{2} \):
\[ x_1 = 2.0 \times -0.7071 \approx -1.414 \text{ cm} \]

2. Substitute \( t = 0.0125 \) into \( x_2 \):
\[ x_2 = 2.0 \sin(120 \pi \times 0.0125 + \frac{\pi}{3}) = 2.0 \sin(1.5 \pi + \frac{\pi}{3}) \]
This simplifies to:
\[ 1.5 \pi + \frac{\pi}{3} = \frac{4.5 \pi + \pi}{3} = \frac{5.5 \pi}{3} \]
Now calculate \( \sin\left(\frac{5.5 \pi}{3}\right) \):
\[ \sin\left(\frac{5.5 \pi}{3}\right) = -\frac{\sqrt{3}}{2} \quad (\text{since } \frac{5.5 \pi}{3} \text{ is in the fourth quadrant}) \]
Thus,
\[ x_2 = 2.0 \times -0.866 \approx -1.732 \text{ cm} \]

3. Total displacement \( x \) at \( t = 0.0125 \):
\[ x = x_1 + x_2 = -1.414 + (-1.732) = -3.146 \text{ cm} \]

**Step 2: Calculate displacement at \( t = 0.025 \) seconds.**

1. Substitute \( t = 0.025 \) into \( x_1 \):
\[ x_1 = 2.0 \sin(100 \pi \times 0.025) = 2.0 \sin(2.5 \pi) \]
Since \( \sin(2.5 \pi) = 1 \):
\[ x_1 = 2.0 \times 1 = 2.0 \text{ cm} \]

2. Substitute \( t = 0.025 \) into \( x_2 \):
\[ x_2 = 2.0 \sin(120 \pi \times 0.025 + \frac{\pi}{3}) = 2.0 \sin(3 \pi + \frac{\pi}{3}) = 2.0 \sin(3 \pi + \frac{\pi}{3}) \]
This simplifies to:
\[ 3 \pi + \frac{\pi}{3} = \frac{9 \pi + \pi}{3} = \frac{10 \pi}{3} \]
Now calculate \( \sin\left(\frac{10 \pi}{3}\right) \):
\[ \sin\left(\frac{10 \pi}{3}\right) = \sin\left(\frac{10 \pi}{3} - 3 \cdot 2\pi\right) = \sin\left(\frac{10 \pi - 18 \pi}{3}\right) = \sin\left(-\frac{8 \pi}{3}\right) = -\frac{\sqrt{3}}{2} \]
Thus,
\[ x_2 = 2.0 \times -0.866 \approx -1.732 \text{ cm} \]

3. Total displacement \( x \) at \( t = 0.025 \):
\[ x = x_1 + x_2 = 2.0 + (-1.732) = 0.268 \text{ cm} \]

### Final Answers:
- Displacement at \( t = 0.0125 \) seconds: \( -3.146 \) cm
- Displacement at \( t = 0.025 \) seconds: \( 0.268 \) cm