Position-time relationship of a particle executing simple harmonic motion is given by equation
`x=2sin(50pit+(2pi)/(3))` where x is in meters and time t is in seconds.
What is the position of particle at t=0.5s ?

Options

A. `sqrt(2)m`
B. `sqrt(3)m`
C. `-sqrt(3)m`
D. 2 m

Answer

To find the position of the particle at \( t = 0.5 \) seconds given the equation of motion \( x = 2 \sin(50\pi t + \frac{2\pi}{3}) \), we will follow these steps:

### Step 1: Substitute the time value into the equation
We need to substitute \( t = 0.5 \) seconds into the position equation.

\[ x = 2 \sin(50\pi(0.5) + \frac{2\pi}{3}) \]

### Step 2: Calculate the argument of the sine function
Now, calculate \( 50\pi(0.5) \):

\[ 50\pi(0.5) = 25\pi \]

Now substitute this back into the equation:

\[ x = 2 \sin(25\pi + \frac{2\pi}{3}) \]

### Step 3: Simplify the sine function
Next, we need to simplify \( 25\pi + \frac{2\pi}{3} \).

To do this, we can express \( 25\pi \) in terms of a common denominator with \( \frac{2\pi}{3} \):

\[ 25\pi = \frac{75\pi}{3} \]

Now, add:

\[ \frac{75\pi}{3} + \frac{2\pi}{3} = \frac{77\pi}{3} \]

### Step 4: Use the periodic property of sine
The sine function has a period of \( 2\pi \), so we can reduce \( \frac{77\pi}{3} \) modulo \( 2\pi \):

\[ \frac{77\pi}{3} \mod 2\pi = \frac{77\pi}{3} - 2\pi \cdot 12 = \frac{77\pi}{3} - \frac{72\pi}{3} = \frac{5\pi}{3} \]

### Step 5: Calculate the sine value
Now we need to find \( \sin\left(\frac{5\pi}{3}\right) \). The angle \( \frac{5\pi}{3} \) is in the fourth quadrant, where sine is negative:

\[ \sin\left(\frac{5\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \]

### Step 6: Substitute back into the equation
Now substitute this value back into the equation for \( x \):

\[ x = 2 \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3} \]

### Conclusion
Thus, the position of the particle at \( t = 0.5 \) seconds is:

\[ \boxed{-\sqrt{3}} \text{ meters} \]