A simple pendulum of length `l` is suspended in a car that is travelling with a constant speed `v` around a circle of radius `r`. If the pendulum undergoes small oscillations about its equilibrium position , what will its freqeuency of oscillation be ?

Options

A. (a)`(1)/(2pi)sqrt((g)/(l))`
B. (b)`(1)/(2pi)sqrt((g)/(R))`
C. (c)`(1)/(2pi)sqrt((g^(2)+(v^(4))/(R^(2)))^(1//2)/(l)`
D. (d)`(1)/(2pi)sqrt((v^(2))/(Rl))`

Explanation

To solve the problem of finding the frequency of oscillation of a simple pendulum suspended in a car traveling in a circular path, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Pendulum:**
- The pendulum experiences two forces:
- The gravitational force acting downward, which is \( g \).
- The centrifugal force due to the circular motion of the car, which can be expressed as \( \frac{v^2}{r} \), acting horizontally towards the center of the circular path.

2. **Determine the Effective Gravity:**
- The effective gravitational acceleration (\( g_{\text{effective}} \)) acting on the pendulum can be found by considering both the gravitational force and the centrifugal force.
- The effective gravity can be calculated using the Pythagorean theorem:
\[ g_{\text{effective}} = \sqrt{g^2 + \left(\frac{v^2}{r}\right)^2} \]

3. **Write the Formula for the Time Period:**
- The time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{l}{g_{\text{effective}}}} \]
- Substituting \( g_{\text{effective}} \) into this formula gives:
\[ T = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + \left(\frac{v^2}{r}\right)^2}}} \]

4. **Calculate the Frequency:**
- The frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g_{\text{effective}}}{l}} \]
- Substituting \( g_{\text{effective}} \) into the frequency formula gives:
\[ f = \frac{1}{2\pi} \sqrt{\frac{\sqrt{g^2 + \left(\frac{v^2}{r}\right)^2}}{l}} \]

5. **Final Expression:**
- Thus, the frequency of oscillation of the pendulum is:
\[ f = \frac{1}{2\pi} \sqrt{\frac{g^2 + \left(\frac{v^2}{r}\right)^2}{l}} \]