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 determining the correct graph for the relationship between acceleration \( a \) and linear velocity \( v \) for a mass \( M \) performing linear simple harmonic motion, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Displacement Equation**:
The displacement \( x \) of a mass in simple harmonic motion can be expressed as:
\[ x = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

**Hint**: Remember that the sine function oscillates between -1 and 1, which is crucial for understanding the motion.

2. **Calculate the Velocity**:
To find the velocity \( v \), we differentiate the displacement with respect to time \( t \):
\[ v = \frac{dx}{dt} = A \omega \cos(\omega t) \]

**Hint**: The velocity is maximum when \( \cos(\omega t) \) is at its maximum value (1).

3. **Calculate the Acceleration**:
The acceleration \( a \) is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t) \]

**Hint**: Notice that acceleration is directly related to the sine function, which indicates its oscillatory nature.

4. **Relate \( v \) and \( a \)**:
From the equations for \( v \) and \( a \), we can express \( \cos(\omega t) \) and \( \sin(\omega t) \):
\[ \cos(\omega t) = \frac{v}{A \omega} \]
\[ \sin(\omega t) = -\frac{a}{A \omega^2} \]

5. **Square Both Equations**:
Squaring both equations gives:
\[ \cos^2(\omega t) = \frac{v^2}{A^2 \omega^2} \]
\[ \sin^2(\omega t) = \frac{a^2}{A^2 \omega^4} \]

6. **Use the Pythagorean Identity**:
Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ \frac{v^2}{A^2 \omega^2} + \frac{a^2}{A^2 \omega^4} = 1 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ v^2 = A^2 \omega^2 - \frac{a^2}{\omega^2} \]
This shows a linear relationship between \( v^2 \) and \( a^2 \).

8. **Identify the Graph**:
The equation \( v^2 = A^2 \omega^2 - \frac{a^2}{\omega^2} \) indicates that as \( a \) increases, \( v^2 \) decreases, which corresponds to a straight line with a negative slope. The x-intercept is positive.

**Hint**: Look for a graph that shows a linear relationship with a negative slope.

9. **Conclusion**:
Based on the analysis, the correct graph that represents the relationship between acceleration \( a \) and linear velocity \( v \) is option **D**.