If force on a particle `vecF = (Sin at)hati + (cos at)hatj` and displacement `vecS = sin(frac{at}{3}hati) + cos(frac{at}{3})hatj` are functions of time (t) then value of t at which they are perpendicular for first time is (a is positive constant and t>0)

Options

A. `t = frac{pi}{2]`
B. `t=frac{3pi}{2a}`
C. `t = frac{3pi}{4a}`
D. `t = frac{pi}{3}`

Answer

To solve the problem, we need to determine the time \( t \) at which the force vector \( \vec{F} \) and the displacement vector \( \vec{S} \) are perpendicular. This occurs when their dot product is zero.

### Step-by-Step Solution:

1. **Identify the Force and Displacement Vectors:**
\[ \vec{F} = \sin(at) \hat{i} + \cos(at) \hat{j} \]
\[ \vec{S} = \sin\left(\frac{at}{3}\right) \hat{i} + \cos\left(\frac{at}{3}\right) \hat{j} \]

2. **Set Up the Dot Product:**
The dot product \( \vec{F} \cdot \vec{S} \) is given by:
\[ \vec{F} \cdot \vec{S} = \left(\sin(at) \hat{i} + \cos(at) \hat{j}\right) \cdot \left(\sin\left(\frac{at}{3}\right) \hat{i} + \cos\left(\frac{at}{3}\right) \hat{j}\right) \]
This simplifies to:
\[ \vec{F} \cdot \vec{S} = \sin(at) \sin\left(\frac{at}{3}\right) + \cos(at) \cos\left(\frac{at}{3}\right) \]

3. **Use the Cosine Angle Identity:**
We can use the cosine angle subtraction identity:
\[ \cos(a - b) = \cos a \cos b + \sin a \sin b \]
Thus, we can rewrite the dot product as:
\[ \vec{F} \cdot \vec{S} = \cos\left(at - \frac{at}{3}\right) = \cos\left(\frac{2at}{3}\right) \]

4. **Set the Dot Product to Zero:**
For the vectors to be perpendicular, we set the dot product to zero:
\[ \cos\left(\frac{2at}{3}\right) = 0 \]

5. **Solve for \( t \):**
The cosine function is zero at odd multiples of \( \frac{\pi}{2} \):
\[ \frac{2at}{3} = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
For the first positive solution (taking \( n = 0 \)):
\[ \frac{2at}{3} = \frac{\pi}{2} \]
Rearranging gives:
\[ t = \frac{3\pi}{4a} \]

### Final Answer:
The value of \( t \) at which the force and displacement vectors are perpendicular for the first time is:
\[ t = \frac{3\pi}{4a} \]