The speed v of a particle moving along a straight line. When it is at distance x from a fixed point on the line is `v^(2)=144-9x^(2)`. Select the correct alternatives

Options

A. The motioni of the particle is SHM with time period `T=(2pi)/3` units
B. The maximum displacement of the particle from the fixed point is 4 units
C. The magnitude of acceleration at a distance 3 units from the fixed point is 27 units
D. the motion of the particles is periodic but not simple harmonic

Answer

To solve the problem, we start with the equation given for the speed \( v \) of a particle:

\[ v^2 = 144 - 9x^2 \]

### Step 1: Differentiate the equation with respect to \( x \)

We differentiate both sides of the equation with respect to \( x \):

\[ \frac{d}{dx}(v^2) = \frac{d}{dx}(144 - 9x^2) \]

### Step 2: Apply the chain rule

Using the chain rule on the left side and differentiating the right side:

\[ 2v \frac{dv}{dx} = 0 - 9 \cdot 2x \]

This simplifies to:

\[ 2v \frac{dv}{dx} = -18x \]

### Step 3: Rearranging the equation

We can rearrange this to express acceleration \( a \):

\[ v \frac{dv}{dx} = -9x \]

### Step 4: Identify the type of motion

Since acceleration is proportional to displacement and opposite in direction, this indicates that the motion is simple harmonic motion (SHM).

### Step 5: Compare with SHM equation

In SHM, the acceleration \( a \) can be expressed as:

\[ a = -\omega^2 x \]

From our equation, we can see that:

\[ \omega^2 = 9 \implies \omega = 3 \]

### Step 6: Calculate the time period \( T \)

The time period \( T \) of SHM is given by:

\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{3} \]

### Step 7: Find the amplitude

To find the amplitude \( A \), we compare the original equation with the standard form of SHM:

\[ v^2 = \omega^2 (A^2 - x^2) \]

From \( v^2 = 144 - 9x^2 \), we can identify:

\[ \omega^2 A^2 = 144 \quad \text{and} \quad \omega^2 = 9 \]

Thus, we have:

\[ A^2 = \frac{144}{9} = 16 \implies A = 4 \]

### Step 8: Calculate the acceleration at \( x = 3 \)

The magnitude of acceleration \( a \) at \( x = 3 \) is given by:

\[ a = -\omega^2 x \]

Substituting \( \omega^2 = 9 \) and \( x = 3 \):

\[ a = -9 \cdot 3 = -27 \]

The magnitude is \( 27 \).

### Conclusion

The correct alternatives based on our calculations are:

1. The motion is SHM.
2. The amplitude is \( 4 \) units.
3. The magnitude of acceleration at \( 3 \) units from the fixed point is \( 27 \) units.

### Summary of Correct Options

- Option 1: Correct (The motion is SHM)
- Option 2: Correct (Amplitude is \( 4 \) units)
- Option 3: Correct (Acceleration at \( 3 \) units is \( 27 \) units)
- Option 4: Incorrect (It is SHM)