The simple harmonic motion of a particle is given by x = a sin `2 pit` . Then, the location of the particle from its mean position at a time `1//8^(th)` of second is

Options

A. a
B. `a/2`
C. `a/sqrt2`
D. `a/4`

Answer

To solve the problem step by step, we will analyze the given equation of motion and substitute the required values to find the location of the particle from its mean position.

### Step-by-Step Solution:

1. **Identify the Equation of Motion**:
The equation of simple harmonic motion (SHM) is given as:
\[ x = a \sin(2 \pi t) \]
where \( a \) is the amplitude and \( t \) is the time.

2. **Determine the Time Period**:
The angular frequency \( \omega \) is given by:
\[ \omega = 2 \pi \]
The time period \( T \) can be calculated using the formula:
\[ T = \frac{2 \pi}{\omega} = \frac{2 \pi}{2 \pi} = 1 \text{ second} \]

3. **Calculate the Time for the Given Condition**:
We need to find the position of the particle at \( \frac{1}{8} \) of a second. Since the time period is 1 second, \( \frac{1}{8} \) of the time period is:
\[ t = \frac{1}{8} \text{ seconds} \]

4. **Substitute the Time into the Equation**:
Now, substitute \( t = \frac{1}{8} \) into the equation:
\[ x = a \sin(2 \pi \cdot \frac{1}{8}) \]

5. **Simplify the Sine Function**:
Calculate the argument of the sine function:
\[ 2 \pi \cdot \frac{1}{8} = \frac{2 \pi}{8} = \frac{\pi}{4} \]
Therefore, the equation becomes:
\[ x = a \sin\left(\frac{\pi}{4}\right) \]

6. **Evaluate the Sine Value**:
The value of \( \sin\left(\frac{\pi}{4}\right) \) is:
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Hence, substituting this value gives:
\[ x = a \cdot \frac{1}{\sqrt{2}} = \frac{a}{\sqrt{2}} \]

7. **Conclusion**:
The location of the particle from its mean position at \( \frac{1}{8} \) of a second is:
\[ x = \frac{a}{\sqrt{2}} \]