A simple pendulum of length 1m is attached to the ceiling of an elevator which is accelerating upward at the rate of `1m//s^(2)`. Its frequency is approximately :-

Options

A. 2 Hz
B. 1.5 Hz
C. 5 Hz
D. 0.5 Hz

Answer

To solve the problem, we need to determine the frequency of a simple pendulum attached to an elevator that is accelerating upwards. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Effective Gravity
When the elevator is accelerating upwards, the effective acceleration due to gravity (G_eff) acting on the pendulum bob is the sum of the gravitational acceleration (g) and the elevator's acceleration (a).

\[ G_{\text{eff}} = g + a \]

Given:
- \( g = 10 \, \text{m/s}^2 \) (approximately)
- \( a = 1 \, \text{m/s}^2 \)

### Step 2: Calculate the Effective Gravity
Now, substituting the values into the equation for effective gravity:

\[ G_{\text{eff}} = 10 \, \text{m/s}^2 + 1 \, \text{m/s}^2 = 11 \, \text{m/s}^2 \]

### Step 3: Determine the Time Period of the Pendulum
The formula for the time period (T) of a simple pendulum is given by:

\[ T = 2\pi \sqrt{\frac{L}{G_{\text{eff}}}} \]

Where:
- \( L = 1 \, \text{m} \) (length of the pendulum)

Substituting the values we have:

\[ T = 2\pi \sqrt{\frac{1}{11}} \]

### Step 4: Calculate the Time Period
Calculating the square root and multiplying by \( 2\pi \):

\[ T = 2\pi \sqrt{\frac{1}{11}} \approx 2\pi \times 0.301 = 1.89 \, \text{s} \]

### Step 5: Calculate the Frequency
The frequency (f) is the reciprocal of the time period:

\[ f = \frac{1}{T} \]

Substituting the value of T:

\[ f = \frac{1}{1.89} \approx 0.529 \, \text{Hz} \]

### Step 6: Round to the Nearest Option
Among the given options, the approximate frequency is:

\[ f \approx 0.5 \, \text{Hz} \]

Thus, the correct answer is option 4.

### Summary of the Solution
The frequency of the simple pendulum in the accelerating elevator is approximately \( 0.5 \, \text{Hz} \).

---