Variation of acceleration a of a particle executing SHM with displacement x is

Options

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Answer

To solve the question regarding the variation of acceleration \( a \) of a particle executing Simple Harmonic Motion (SHM) with displacement \( x \), we can follow these steps:

### Step 1: Understand the relationship in SHM
In SHM, the displacement \( x \) of a particle can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

### Step 2: Derive the expression for acceleration
The acceleration \( a \) of the particle is given by the second derivative of displacement with respect to time:
\[ a = \frac{d^2x}{dt^2} \]
To find \( a \), we first differentiate \( x(t) \):
\[ v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]
Now, differentiate \( v \) to find \( a \):
\[ a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) \]

### Step 3: Relate acceleration to displacement
Using the original displacement equation, we can express acceleration in terms of displacement \( x \):
\[ a = -\omega^2 x \]
This shows that acceleration \( a \) is directly proportional to displacement \( x \) but in the opposite direction (hence the negative sign).

### Step 4: Graphical representation
The equation \( a = -\omega^2 x \) represents a straight line through the origin with a negative slope of \( -\omega^2 \). This means that as \( x \) increases, \( a \) decreases, and vice versa.

### Conclusion
The variation of acceleration \( a \) with displacement \( x \) is a straight line passing through the origin with a negative slope.

### Final Answer
The correct option is a straight line passing through the origin with a negative slope.

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