A simple pendulum performs simple harmonic motion about x=0 with an amplitude a and time period T. The speed of the pendulum at x=a/2 will be:

• A. πa/T
• B. 3π²a/T
• C. πa√3/T
• D. πa√3/2T

To find the speed of a simple pendulum at the position \( x = \frac{a}{2} \), we can use the principles of simple harmonic motion (SHM).

Key Concepts

• Amplitude (a): The maximum displacement from the equilibrium position.
• Time Period (T): The time taken for one complete cycle of motion.
• Speed in SHM: The speed at any position can be calculated using the formula:

Speed Formula

The speed \( v \) at a displacement \( x \) in SHM is given by:

v = ω√(a² - x²)

where \( ω \) (angular frequency) is defined as:

ω = 2π/T

Calculating Speed at \( x = \frac{a}{2} \)

Substituting \( x = \frac{a}{2} \) into the speed formula:

v = ω√(a² - (a/2)²)

Now, calculate \( a² - (a/2)² \):

a² - (a²/4) = (4a²/4) - (a²/4) = (3a²/4)

Thus, we have:

v = ω√(3a²/4) = ω(a√3/2)

Substituting for \( ω \)

Now, substituting \( ω = 2π/T \):

v = (2π/T)(a√3/2) = (πa√3/T)

Final Answer

The speed of the pendulum at \( x = \frac{a}{2} \) is:

πa√3/T

Therefore, the correct option is C. πa√3/T.