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

Options

A. `(pi a sqrt(3))/(2T)`
B. `(pi a)/(T)`
C. `(3pi^(2) a)/(T)`
D. `(pi a sqrt(3))/(T)`

Answer

To find the speed of a simple pendulum at a position \( x = \frac{a}{2} \), we can use the principles of simple harmonic motion (SHM). Here’s a step-by-step solution:

### Step 1: Understand the motion of the pendulum
The motion of a simple pendulum can be described by the equation of SHM:
\[ x(t) = a \sin(\omega t) \]
where:
- \( x(t) \) is the displacement from the mean position,
- \( a \) is the amplitude,
- \( \omega \) is the angular frequency.

### Step 2: Find the expression for velocity
The velocity \( v(t) \) of the pendulum can be derived from the displacement:
\[ v(t) = \frac{dx}{dt} = a \omega \cos(\omega t) \]
Alternatively, we can express the velocity in terms of displacement \( x \):
\[ v = \omega \sqrt{a^2 - x^2} \]

### Step 3: Substitute \( x = \frac{a}{2} \)
Now, we need to find the velocity when \( x = \frac{a}{2} \):
\[ v = \omega \sqrt{a^2 - \left(\frac{a}{2}\right)^2} \]

### Step 4: Simplify the expression
Calculate \( a^2 - \left(\frac{a}{2}\right)^2 \):
\[ \left(\frac{a}{2}\right)^2 = \frac{a^2}{4} \]
Thus:
\[ a^2 - \frac{a^2}{4} = \frac{4a^2}{4} - \frac{a^2}{4} = \frac{3a^2}{4} \]
Now substitute this back into the velocity equation:
\[ v = \omega \sqrt{\frac{3a^2}{4}} = \omega \frac{a \sqrt{3}}{2} \]

### Step 5: Relate \( \omega \) to the time period \( T \)
The angular frequency \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the velocity equation gives:
\[ v = \frac{2\pi}{T} \cdot \frac{a \sqrt{3}}{2} = \frac{\pi a \sqrt{3}}{T} \]

### Final Result
Thus, the speed of the pendulum at \( x = \frac{a}{2} \) is:
\[ v = \frac{\pi a \sqrt{3}}{T} \]

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