The equation of motion of a particle executing SHM is `((d^2 x)/(dt^2))+kx=0`. The time period of the particle will be :

Options

A. `2pi // sqrtk`
B. `2pi // k`
C. `2pik`
D. `2pi sqrtk`

Answer

To find the time period of a particle executing simple harmonic motion (SHM) given the equation of motion \( \frac{d^2 x}{dt^2} + kx = 0 \), we can follow these steps:

### Step 1: Understand the Equation of Motion
The equation given is a standard form of the equation of motion for SHM. It can be rewritten as:
\[ \frac{d^2 x}{dt^2} = -kx \]
This indicates that the acceleration of the particle is proportional to its displacement and directed towards the mean position.

**Hint:** Recognize that this equation describes a restoring force that is characteristic of SHM.

### Step 2: Relate Acceleration to Angular Frequency
From the theory of SHM, we know that the acceleration \( a \) can also be expressed as:
\[ a = \frac{d^2 x}{dt^2} = -\omega^2 x \]
where \( \omega \) is the angular frequency of the motion.

**Hint:** Remember that in SHM, acceleration is always directed towards the mean position and is proportional to the displacement.

### Step 3: Compare the Two Expressions for Acceleration
From the two expressions for acceleration, we can equate:
\[ -kx = -\omega^2 x \]
This leads to:
\[ k = \omega^2 \]

**Hint:** This step shows how the constant \( k \) in the equation relates to the angular frequency \( \omega \).

### Step 4: Find the Angular Frequency
From the relation \( k = \omega^2 \), we can express \( \omega \) as:
\[ \omega = \sqrt{k} \]

**Hint:** The angular frequency \( \omega \) is a key parameter in determining the time period.

### Step 5: Calculate the Time Period
The time period \( T \) of the SHM is given by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \omega = \sqrt{k} \) into this formula gives:
\[ T = \frac{2\pi}{\sqrt{k}} \]

**Hint:** The time period is inversely proportional to the square root of the spring constant \( k \).

### Final Answer
Thus, the time period of the particle executing SHM is:
\[ T = 2\pi \sqrt{\frac{1}{k}} \]

### Summary
- The time period \( T \) is derived from the relationship between acceleration and displacement in SHM.
- The final expression for the time period is \( T = 2\pi \sqrt{\frac{1}{k}} \).