A particle moves in x-y plane according to the equation `bar(r ) = (hat(i) + 2 hat(j)) A cos omega t` the motion of the particle is

Options

A. on a straight line with a constant speed
B. on an ellipse
C. periodic not SHM
D. simple harmonic

Answer

To determine the type of motion of a particle described by the equation \(\bar{r} = ( \hat{i} + 2 \hat{j}) A \cos(\omega t)\), we can follow these steps:

### Step 1: Understand the given equation
The position vector \(\bar{r}\) of the particle is given as:
\[ \bar{r} = ( \hat{i} + 2 \hat{j}) A \cos(\omega t) \]
This can be rewritten by distributing \(A \cos(\omega t)\):
\[ \bar{r} = A \cos(\omega t) \hat{i} + 2A \cos(\omega t) \hat{j} \]

### Step 2: Identify the components of motion
From the rewritten equation, we can identify the x and y components of the motion:
- \(x = A \cos(\omega t)\)
- \(y = 2A \cos(\omega t)\)

### Step 3: Analyze the equations for x and y
Both \(x\) and \(y\) are functions of time \(t\) and are expressed as cosine functions:
- \(x = A \cos(\omega t)\)
- \(y = 2A \cos(\omega t)\)

### Step 4: Check for simple harmonic motion (SHM)
In simple harmonic motion, the displacement can be described by equations of the form:
\[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \]
The solutions to this equation are sinusoidal functions (sine or cosine). Since both \(x\) and \(y\) are expressed as cosine functions, they both exhibit SHM.

### Step 5: Conclusion
Since both components of the motion (x and y) are simple harmonic, we conclude that the motion of the particle is simple harmonic motion in both the x and y directions.

### Final Answer
The motion of the particle is **simple harmonic motion**.

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