A particle is performing a linear simple harmonic motion. If the instantaneous acceleration and velocity of the particle are `a` and `v` respectively, identify the graph which correctly represents the relation between `a` and `v`.

Options

A.
B.
C.
D.

Explanation

To solve the problem of identifying the graph that correctly represents the relationship between instantaneous acceleration (a) and velocity (v) of a particle performing simple harmonic motion (SHM), we can follow these steps:

### Step 1: Understand the equations of SHM
In simple harmonic motion, the instantaneous velocity (v) and acceleration (a) can be expressed in terms of the maximum velocity (V₀) and angular frequency (ω). The equations are:
- Velocity: \( v = V_0 \sin(\omega t) \)
- Acceleration: \( a = -V_0 \omega \cos(\omega t) \)

### Step 2: Differentiate the velocity equation
To find the relationship between acceleration and velocity, we can differentiate the velocity equation with respect to time:
- The instantaneous acceleration (a) is given by:
\[ a = \frac{dv}{dt} = V_0 \omega \cos(\omega t) \]

### Step 3: Relate acceleration and velocity
From the equations of SHM, we can express the cosine term in terms of sine:
- We can rewrite the acceleration equation as:
\[ a = -\frac{V_0}{\omega} \sqrt{1 - \left(\frac{v}{V_0}\right)^2} \]
This shows that acceleration (a) is related to velocity (v) through the maximum velocity (V₀).

### Step 4: Square both sides
To eliminate the square root, we can square both sides of the equation:
\[ a^2 = \left(-\frac{V_0}{\omega}\right)^2 \left(1 - \left(\frac{v}{V_0}\right)^2\right) \]
This leads to:
\[ a^2 = \frac{V_0^2}{\omega^2} - \frac{a^2 v^2}{V_0^2} \]

### Step 5: Rearranging the equation
Rearranging gives us a linear relationship:
\[ \frac{v^2}{V_0^2} + \frac{a^2}{\left(\frac{V_0}{\omega}\right)^2} = 1 \]
This can be interpreted as a circle equation in the a-v plane.

### Step 6: Identify the graph
The relationship indicates that as velocity (v) increases, acceleration (a) decreases, and vice versa. The graph of this relationship is a downward-sloping curve. However, when we consider the squares of the variables, we can see that the relationship can also be represented as a straight line in the form:
\[ v^2 + \frac{a^2}{\left(\frac{V_0}{\omega}\right)^2} = constant \]
This indicates a linear relationship with a negative slope.

### Conclusion
From the analysis, we find that the correct graph representing the relationship between acceleration (a) and velocity (v) in simple harmonic motion is a straight line with a negative slope.