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, we need to find the location of the particle from its mean position at a time \( t = \frac{1}{8} \) seconds, given the equation of motion \( x = a \sin(2 \pi t) \).

### Step-by-Step Solution:

1. **Identify the given equation**:
The equation of motion is given as:
\[ x = a \sin(2 \pi t) \]
Here, \( a \) is the amplitude of the motion.

2. **Substitute the time into the equation**:
We need to find the position \( x \) at \( t = \frac{1}{8} \) seconds. Substitute \( t \) into the equation:
\[ x = a \sin(2 \pi \cdot \frac{1}{8}) \]

3. **Calculate the argument of the sine function**:
Simplify the argument:
\[ 2 \pi \cdot \frac{1}{8} = \frac{2\pi}{8} = \frac{\pi}{4} \]

4. **Evaluate the sine function**:
Now, we can evaluate the sine:
\[ x = a \sin\left(\frac{\pi}{4}\right) \]
We know that:
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Therefore:
\[ x = a \cdot \frac{1}{\sqrt{2}} = \frac{a}{\sqrt{2}} \]

5. **Conclusion**:
Thus, the location of the particle from its mean position at \( t = \frac{1}{8} \) seconds is:
\[ x = \frac{a}{\sqrt{2}} \]

### Final Answer:
The location of the particle from its mean position at \( t = \frac{1}{8} \) seconds is \( \frac{a}{\sqrt{2}} \).

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