The equation of motion of a particle is `x=a " cos"(alpha t)^(2)`. The motion is

Options

A. periodic but not simple oscillatory
B. periodic and oscillatory
C. oscillatory but not periodic
D. Neither periodic nor oscillatory

Explanation

To determine the nature of the motion described by the equation \( x = a \cos(\alpha t^2) \), we will analyze the characteristics of the function step by step.

### Step 1: Identify the form of the equation
The given equation is:
\[ x = a \cos(\alpha t^2) \]
This indicates that the displacement \( x \) is a cosine function of \( t^2 \).

**Hint:** Recognize that the cosine function is periodic, but we need to analyze how the argument of the cosine function affects periodicity.

### Step 2: Check for periodicity
A function \( f(t) \) is periodic if there exists a time period \( T \) such that:
\[ f(t) = f(t + T) \]
For our function, we need to check if:
\[ \cos(\alpha t^2) = \cos(\alpha (t + T)^2) \]

Expanding the right-hand side:
\[ \cos(\alpha (t + T)^2) = \cos(\alpha (t^2 + 2tT + T^2)) = \cos(\alpha t^2 + 2\alpha tT + \alpha T^2) \]

**Hint:** Use the property of cosine that states \( \cos A = \cos B \) implies \( A = B + 2n\pi \) for some integer \( n \).

### Step 3: Set up the equation
For the two cosine expressions to be equal, we need:
\[ \alpha t^2 = \alpha t^2 + 2\alpha tT + \alpha T^2 + 2n\pi \]
This simplifies to:
\[ 0 = 2\alpha tT + \alpha T^2 + 2n\pi \]

**Hint:** Rearranging the equation will help us isolate terms related to \( T \).

### Step 4: Analyze the equation
From the equation:
\[ 2\alpha tT + \alpha T^2 + 2n\pi = 0 \]
This indicates that for different values of \( t \), the left side will yield different results unless \( T \) is zero, which is not a valid period. Thus, there is no single \( T \) that satisfies the equation for all \( t \).

**Hint:** Consider the implications of having a variable \( t \) in the equation.

### Step 5: Conclusion
Since we cannot find a constant period \( T \) that satisfies the periodic condition for all \( t \), the motion described by \( x = a \cos(\alpha t^2) \) is not periodic. However, it is oscillatory because it involves a cosine function.

Thus, we conclude that the motion is oscillatory but not periodic.

**Final Answer:** The motion is oscillatory but not periodic.