A simple pendulum of length 1 feet suspended from the ceiling of an elevator takes `pi/3` seconds to complete one oscilation. Find the acceleration of the elevator.

Explanation

To solve the problem of finding the acceleration of the elevator in which a simple pendulum of length 1 foot takes \(\frac{\pi}{3}\) seconds to complete one oscillation, we can follow these steps:

### Step 1: Understand the Effective Gravity
When the elevator accelerates upwards with acceleration \(a\), the effective gravitational acceleration \(g_{\text{effective}}\) inside the elevator is given by:
\[ g_{\text{effective}} = g + a \]
where \(g\) is the acceleration due to gravity. In feet per second squared, \(g\) is approximately \(32 \, \text{ft/s}^2\).

### Step 2: Use the Formula for the Time Period of a Pendulum
The time period \(T\) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
where \(L\) is the length of the pendulum.

### Step 3: Substitute Known Values
Given that \(L = 1 \, \text{ft}\) and \(T = \frac{\pi}{3} \, \text{s}\), we can substitute these values into the time period formula:
\[ \frac{\pi}{3} = 2\pi \sqrt{\frac{1}{g + a}} \]

### Step 4: Simplify the Equation
Dividing both sides by \(2\pi\):
\[ \frac{1}{6} = \sqrt{\frac{1}{g + a}} \]

### Step 5: Square Both Sides
Squaring both sides to eliminate the square root gives:
\[ \left(\frac{1}{6}\right)^2 = \frac{1}{g + a} \]
\[ \frac{1}{36} = \frac{1}{g + a} \]

### Step 6: Rearrange to Solve for \(g + a\)
Taking the reciprocal of both sides:
\[ g + a = 36 \]

### Step 7: Substitute the Value of \(g\)
Substituting \(g = 32 \, \text{ft/s}^2\):
\[ 32 + a = 36 \]

### Step 8: Solve for \(a\)
Now, solving for \(a\):
\[ a = 36 - 32 = 4 \, \text{ft/s}^2 \]

### Final Answer
The acceleration of the elevator is \(4 \, \text{ft/s}^2\).

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