A simple pendulum of length l is suspended from the ceilling of a car moving with a speed v on a circular horizontal road of radius r. a. Find the tension in the string when it is at rest with respect to the car. b.Find the time period of small oscillation.

Explanation

To solve the problem step by step, we will address both parts: finding the tension in the string and the time period of small oscillations.

### Part (a): Finding the Tension in the String

1. **Identify Forces Acting on the Pendulum Bob:**
- The forces acting on the pendulum bob are:
- The gravitational force \( mg \) acting downward.
- The tension \( T \) in the string acting along the string.
- The centrifugal force \( F_c \) acting outward due to the circular motion of the car, given by \( F_c = \frac{mv^2}{r} \).

2. **Draw the Free Body Diagram (FBD):**
- The tension can be resolved into two components:
- Vertical component: \( T \cos \theta = mg \)
- Horizontal component: \( T \sin \theta = \frac{mv^2}{r} \)

3. **Square and Add the Two Equations:**
- From the two equations, we can square both sides:
\[ (T \cos \theta)^2 = (mg)^2 \quad \text{(1)} \]
\[ (T \sin \theta)^2 = \left(\frac{mv^2}{r}\right)^2 \quad \text{(2)} \]
- Adding (1) and (2):
\[ T^2 \cos^2 \theta + T^2 \sin^2 \theta = m^2 g^2 + \left(\frac{mv^2}{r}\right)^2 \]
- Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ T^2 = m^2 g^2 + \left(\frac{mv^2}{r}\right)^2 \]

4. **Solve for Tension \( T \):**
- Taking the square root:
\[ T = \sqrt{m^2 g^2 + \left(\frac{mv^2}{r}\right)^2} \]
- Factoring out \( m \):
\[ T = m \sqrt{g^2 + \left(\frac{v^2}{r}\right)^2} \]

### Part (b): Finding the Time Period of Small Oscillations

1. **Effective Gravitational Acceleration:**
- The effective gravitational acceleration \( g_{\text{effective}} \) in the car's frame is given by:
\[ g_{\text{effective}} = \sqrt{g^2 + \left(\frac{v^2}{r}\right)^2} \]

2. **Time Period Formula:**
- The time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{l}{g_{\text{effective}}}} \]

3. **Substituting \( g_{\text{effective}} \):**
- Substitute \( g_{\text{effective}} \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + \left(\frac{v^2}{r}\right)^2}}} \]
- This simplifies to:
\[ T = 2\pi \frac{l^{1/2}}{(g^2 + \left(\frac{v^2}{r}\right)^2)^{1/4}} \]

### Final Answers:
- **Tension in the String:**
\[ T = m \sqrt{g^2 + \left(\frac{v^2}{r}\right)^2} \]
- **Time Period of Small Oscillations:**
\[ T = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + \left(\frac{v^2}{r}\right)^2}}} \]