The distance between the point of suspension and the centre of gravity of a compound pendulum is `l` and the radius of gyration about the horizontal axis through the centre of gravity is `k`, then its time period will be

Options

A. `2pisqrt((l+k)/(g))`
B. `2pisqrt((l^(2)+k^(2))/(lg))`
C. `2pisqrt((l+k^(2))/(g))`
D. `2pisqrt((2k)/(lg))`

Explanation

To find the time period of a compound pendulum, we can follow these steps:

### Step 1: Understand the Setup
In a compound pendulum, the distance between the point of suspension and the center of gravity is denoted as \( l \), and the radius of gyration about the horizontal axis through the center of gravity is denoted as \( k \).

### Step 2: Calculate the Torque
The torque (\( \tau \)) due to the weight of the pendulum at the center of gravity can be expressed as:
\[ \tau = m g l \sin(\theta) \]
where \( m \) is the mass of the pendulum, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of displacement.

### Step 3: Small Angle Approximation
For small angles, we can approximate \( \sin(\theta) \approx \theta \). Thus, the torque becomes:
\[ \tau \approx m g l \theta \]

### Step 4: Moment of Inertia
The moment of inertia (\( I \)) about the point of suspension can be calculated using the parallel axis theorem:
\[ I = I_c + m l^2 \]
where \( I_c \) is the moment of inertia about the center of gravity, which can be expressed as:
\[ I_c = m k^2 \]
Thus, the total moment of inertia becomes:
\[ I = m k^2 + m l^2 = m (k^2 + l^2) \]

### Step 5: Relate Torque to Angular Acceleration
The torque can also be expressed in terms of angular acceleration (\( \alpha \)):
\[ \tau = I \alpha \]
Substituting the expression for torque and moment of inertia, we have:
\[ m g l \theta = (m (k^2 + l^2)) \alpha \]

### Step 6: Cancel Mass and Rearrange
Cancel \( m \) from both sides:
\[ g l \theta = (k^2 + l^2) \alpha \]
Since \( \alpha = \frac{d^2\theta}{dt^2} \), we can write:
\[ g l \theta = (k^2 + l^2) \frac{d^2\theta}{dt^2} \]

### Step 7: Identify the Angular Frequency
This is a standard form of the equation of motion for simple harmonic motion, which can be rearranged to:
\[ \frac{d^2\theta}{dt^2} + \frac{g l}{k^2 + l^2} \theta = 0 \]
From this, we can identify the angular frequency \( \omega^2 \) as:
\[ \omega^2 = \frac{g l}{k^2 + l^2} \]

### Step 8: Relate Angular Frequency to Time Period
The relationship between angular frequency and time period \( T \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Thus, we can express the time period \( T \) as:
\[ T = 2\pi \sqrt{\frac{k^2 + l^2}{g l}} \]

### Final Answer
The time period \( T \) of the compound pendulum is:
\[ T = 2\pi \sqrt{\frac{k^2 + l^2}{g l}} \]

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