A mass M is performing linear simple harmonic motion. Then correct graph for acceleration a and corresponding linear velocity v is

Options

A.
B.
C.
D.

Answer

To solve the problem of identifying the correct graph for acceleration \( a \) and corresponding linear velocity \( v \) of a mass performing linear simple harmonic motion (SHM), we can follow these steps:

### Step 1: Understand the equations of SHM
The displacement \( x \) of a mass in SHM can be expressed as:
\[ x(t) = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

### Step 2: Derive the velocity equation
To find the velocity \( v \), we differentiate the displacement with respect to time:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t) \]

### Step 3: Derive the acceleration equation
Next, we differentiate the velocity to find the acceleration \( a \):
\[ a(t) = \frac{dv}{dt} = -A \omega^2 \sin(\omega t) \]
This can also be expressed in terms of displacement:
\[ a = -\omega^2 x \]

### Step 4: Relate acceleration and velocity
From the equations derived, we have:
- \( v = A \omega \cos(\omega t) \)
- \( a = -\omega^2 A \sin(\omega t) \)

### Step 5: Express \( v^2 \) and \( a^2 \)
From the equations, we can express \( v^2 \) and \( a^2 \):
\[ v^2 = (A \omega \cos(\omega t))^2 = A^2 \omega^2 \cos^2(\omega t) \]
\[ a^2 = (-\omega^2 A \sin(\omega t))^2 = \omega^4 A^2 \sin^2(\omega t) \]

### Step 6: Use the identity \( \sin^2 + \cos^2 = 1 \)
Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we can relate \( v^2 \) and \( a^2 \):
\[ \frac{v^2}{A^2 \omega^2} + \frac{a^2}{\omega^4 A^2} = 1 \]
Rearranging gives:
\[ v^2 = A^2 \omega^2 - \frac{a^2}{\omega^4} A^2 \omega^2 \]
This shows that \( v^2 \) is a linear function of \( a^2 \) with a negative slope.

### Step 7: Identify the graph
The relationship derived indicates that as \( a^2 \) increases, \( v^2 \) decreases, which corresponds to a straight line with a negative slope. Therefore, the correct graph will show \( v^2 \) on the y-axis and \( a^2 \) on the x-axis, with a negative slope.

### Conclusion
Based on the analysis, the correct option for the graph of acceleration \( a \) and corresponding linear velocity \( v \) is the one that shows a linear relationship with a negative slope.

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