The time period of a simple pendulum of length L as measured in an elevator descending with acceleration g / 3 is

Options

A. `2pisqrt((3L)/(g))`
B. `pisqrt(((3L)/(g)))`
C. `2pisqrt(((3L)/(2g)))`
D. `2pisqrt(((3L)/(3g)))`

Explanation

To find the time period of a simple pendulum of length \( L \) in an elevator descending with an acceleration of \( \frac{g}{3} \), we can follow these steps:

### Step 1: Determine the effective gravitational acceleration
When the elevator is descending with an acceleration of \( \frac{g}{3} \), the effective gravitational acceleration \( g_{\text{effective}} \) acting on the pendulum can be calculated as follows:

\[ g_{\text{effective}} = g - a \]

where \( a \) is the acceleration of the elevator. Here, \( a = \frac{g}{3} \).

Substituting this into the equation gives:

\[ g_{\text{effective}} = g - \frac{g}{3} = g \left(1 - \frac{1}{3}\right) = g \left(\frac{2}{3}\right) = \frac{2g}{3} \]

### Step 2: Write the formula for the time period of a simple pendulum
The time period \( T \) of a simple pendulum is given by the formula:

\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]

### Step 3: Substitute the effective gravitational acceleration into the formula
Now, substituting \( g_{\text{effective}} = \frac{2g}{3} \) into the time period formula:

\[ T = 2\pi \sqrt{\frac{L}{\frac{2g}{3}}} \]

### Step 4: Simplify the expression
To simplify the expression, we can rewrite it as:

\[ T = 2\pi \sqrt{\frac{3L}{2g}} \]

### Final Result
Thus, the time period of the simple pendulum in an elevator descending with acceleration \( \frac{g}{3} \) is:

\[ T = 2\pi \sqrt{\frac{3L}{2g}} \]