A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line between angular limit `-phi` and `+phi`. For an angular displacement `theta(|theta|ltphi)`, the tension in the string and the velocity of the bob are T and V respectively. The following relations hold good under the above conditions:

Options

A. (a) `Tcos theta=Mg`
B. (b) `T-Mgcostheta=(MV^2)/(L)`
C. (c) The magnitude of the tangential acceleration of the bob `a_T=g sin theta`
D. (d) `T=Mg cos theta`

Explanation

To solve the problem regarding the simple pendulum, we need to analyze the forces acting on the bob when it is displaced by an angle θ (where |θ| < φ).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Bob:**
- The weight of the bob (mg) acts vertically downward.
- The tension (T) in the string acts along the string towards the pivot.

2. **Resolve the Weight into Components:**
- The weight can be resolved into two components:
- A component along the direction of the string: \( mg \cos \theta \)
- A component perpendicular to the string (tangential direction): \( mg \sin \theta \)

3. **Apply Newton's Second Law:**
- In the radial direction (along the string), the net force is given by:
\[ T - mg \cos \theta = \frac{mv^2}{L} \]
- Here, \( \frac{mv^2}{L} \) is the centripetal force required to keep the bob moving in a circular path of radius L.

4. **Tangential Acceleration:**
- The tangential force acting on the bob is \( mg \sin \theta \).
- According to Newton's second law, the tangential acceleration \( a_t \) is given by:
\[ a_t = \frac{F_t}{m} = \frac{mg \sin \theta}{m} = g \sin \theta \]

5. **Summary of Relations:**
- From the above analysis, we have two important relations:
1. \( T - mg \cos \theta = \frac{mv^2}{L} \)
2. \( a_t = g \sin \theta \)

### Final Relations:
- The tension in the string and the velocity of the bob can be expressed as:
- \( T = mg \cos \theta + \frac{mv^2}{L} \)
- \( a_t = g \sin \theta \)