The equation of SHM of a particle is `(d^2y)/(dt^2)+ky=0`, where k is a positive constant. The time period of motion is

Options

A. `(2pi)/(sqrt3)`
B. `(2pi)/(k)`
C. `k/2`
D. `(sqrtk)/(2pi)`

Explanation

To find the time period of motion for a particle undergoing simple harmonic motion (SHM) described by the equation:

\[ \frac{d^2y}{dt^2} + ky = 0 \]

where \( k \) is a positive constant, we can follow these steps:

### Step 1: Identify the form of the equation
The given equation is a second-order linear differential equation that represents SHM. It can be rewritten in the standard form of SHM:

\[ \frac{d^2y}{dt^2} = -ky \]

### Step 2: Relate acceleration to displacement
In SHM, the acceleration \( a \) is proportional to the displacement \( y \) and is directed towards the mean position. This can be expressed as:

\[ a = \frac{d^2y}{dt^2} = -\omega^2 y \]

where \( \omega \) is the angular frequency.

### Step 3: Compare the equations
From the equations:

\[ \frac{d^2y}{dt^2} = -ky \]
and
\[ \frac{d^2y}{dt^2} = -\omega^2 y \]

we can equate the coefficients of \( y \):

\[ \omega^2 = k \]

### Step 4: Solve for angular frequency
Taking the square root of both sides gives us the angular frequency:

\[ \omega = \sqrt{k} \]

### Step 5: Find the time period
The time period \( T \) of SHM is related to the angular frequency \( \omega \) by the formula:

\[ T = \frac{2\pi}{\omega} \]

Substituting \( \omega = \sqrt{k} \) into this equation gives:

\[ T = \frac{2\pi}{\sqrt{k}} \]

### Final Answer
Thus, the time period of the motion is:

\[ T = \frac{2\pi}{\sqrt{k}} \]

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