A simple pendulum is oscillating with an angular amplitude 60°. If the mass of the bob is 50 g, the tension in the string at the mean position is:
Consider: g = 10 m/s², length of the string L = 1 m.
A. 0.5 N
B. 1 N
C. 1.5 N
D. 2 N

We are asked to find the tension in the string of a simple pendulum at the mean position. Given the following data:

Angular amplitude, θ = 60°
Mass of the bob, m = 50g = 0.05kg
Gravitational acceleration, g = 10 m/s²
Length of the string, L = 1m
Step 1: Understanding the forces
At the mean position (the lowest point of the swing), the forces acting on the bob are:

The tension (T) in the string, directed upwards.
The gravitational force (mg), directed downwards.
At the mean position, the tension must balance the gravitational force and provide the necessary centripetal force for the circular motion of the pendulum.

Step 2: Centripetal force
The bob moves in a circular path, so it experiences centripetal acceleration at the mean position. The centripetal force (Fc) required to maintain this motion is given by:

Fc = m * v² / L

Where:

m is the mass of the bob,
v is the velocity of the bob at the mean position,
L is the length of the string.
Step 3: Velocity at the mean position
The bob has maximum velocity at the mean position, where all its potential energy has been converted into kinetic energy. The potential energy at the maximum displacement (at the amplitude of 60°) is:

PE = m * g * h

Where h is the vertical height that the bob has descended. This height can be found using the formula:

h = L * (1 - cosθ)

At the mean position, the kinetic energy (KE) of the bob is equal to the potential energy at the amplitude:

KE = 1/2 * m * v²

Equating the potential and kinetic energy:

m * g * h = 1/2 * m * v²

Simplifying:

g * L * (1 - cosθ) = 1/2 * v²

Solving for v²:

v² = 2 * g * L * (1 - cosθ)

Substituting the values:

v² = 2 * 10 * 1 * (1 - cos60°) v² = 2 * 10 * (1 - 0.5) v² = 2 * 10 * 0.5 v² = 10 m²/s²

Step 4: Centripetal force at the mean position
Now we can find the centripetal force:

Fc = m * v² / L Fc = 0.05 * 10 / 1 Fc = 0.5 N

Step 5: Tension at the mean position
At the mean position, the total tension in the string (T) is the sum of the gravitational force and the centripetal force:

T = mg + Fc T = (0.05 * 10) + 0.5 T = 0.5 + 0.5 T = 1 N

Final Answer:
The tension in the string at the mean position is 1 N. Therefore, the correct answer is:

B. 1 N