A graph of the square of the velocity against the square of the acceleration of a given simple harmonic motion is

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Explanation

To solve the problem of determining the nature of the graph of the square of the velocity against the square of the acceleration in simple harmonic motion (SHM), we can follow these steps:

### Step 1: Understand the equations of SHM
In simple harmonic motion, the displacement \( x \) can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

### Step 2: Find the expression for velocity
The velocity \( v \) is the derivative of displacement with respect to time:
\[ v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]
Thus, the square of the velocity is:
\[ v^2 = (A \omega \cos(\omega t + \phi))^2 = A^2 \omega^2 \cos^2(\omega t + \phi) \]

### Step 3: Find the expression for acceleration
The acceleration \( a \) is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) \]
Thus, the square of the acceleration is:
\[ a^2 = (-A \omega^2 \sin(\omega t + \phi))^2 = A^2 \omega^4 \sin^2(\omega t + \phi) \]

### Step 4: Relate \( v^2 \) and \( a^2 \)
Now we have:
\[ v^2 = A^2 \omega^2 \cos^2(\omega t + \phi) \]
\[ a^2 = A^2 \omega^4 \sin^2(\omega t + \phi) \]

### Step 5: Express \( \sin^2 \) and \( \cos^2 \) in terms of \( v^2 \) and \( a^2 \)
From the identities \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we can express:
\[ \sin^2(\omega t + \phi) = \frac{a^2}{A^2 \omega^4} \]
\[ \cos^2(\omega t + \phi) = \frac{v^2}{A^2 \omega^2} \]

### Step 6: Formulate the relationship
Substituting these into the identity gives:
\[ \frac{a^2}{A^2 \omega^4} + \frac{v^2}{A^2 \omega^2} = 1 \]
Multiplying through by \( A^2 \omega^4 \) yields:
\[ a^2 + \frac{v^2}{\omega^2} = A^2 \omega^4 \]

### Step 7: Identify the graph
This equation represents an ellipse in the \( v^2 \) vs \( a^2 \) plane, where:
- The term \( a^2 \) corresponds to one axis,
- The term \( \frac{v^2}{\omega^2} \) corresponds to the other axis.

### Conclusion
Thus, the graph of the square of the velocity against the square of the acceleration in simple harmonic motion is an ellipse.