For a particle executing SHM along x-axis, force acting on it, is given by `(A, k gt 0)`

Options

A. `-Akx`
B. A cos kx
C. A exp (-kx)
D. Akx

Explanation

To solve the problem of finding the force acting on a particle executing Simple Harmonic Motion (SHM) along the x-axis, we can follow these steps:

### Step 1: Understanding SHM
In SHM, the displacement of the particle can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

### Step 2: Finding Velocity
The velocity \( v(t) \) of the particle is the first derivative of the displacement with respect to time:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]

### Step 3: Finding Acceleration
The acceleration \( a(t) \) is the second derivative of the displacement:
\[ a(t) = \frac{d^2x}{dt^2} = -A \omega^2 \sin(\omega t + \phi) \]
This can also be expressed in terms of displacement \( x \):
\[ a(t) = -\omega^2 x \]

### Step 4: Applying Newton's Second Law
According to Newton's second law, the force \( F \) acting on the particle is given by:
\[ F = m \cdot a \]
Substituting the expression for acceleration:
\[ F = m \cdot (-\omega^2 x) \]
Thus, we have:
\[ F = -m \omega^2 x \]

### Step 5: Identifying the Spring Constant
In SHM, the force can also be expressed in terms of a spring constant \( k \):
\[ F = -k x \]
By comparing the two expressions for force, we can identify that:
\[ k = m \omega^2 \]

### Final Expression
Therefore, the force acting on a particle executing SHM along the x-axis is:
\[ F = -k x \]
where \( k = m \omega^2 \).

### Summary
The force acting on the particle is a restoring force that is directly proportional to the displacement and acts in the opposite direction.

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