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 of finding the frequency of a simple pendulum in an accelerating elevator, we can follow these steps:

### Step 1: Understand the Effective Gravity
When the elevator is accelerating upwards, the effective acceleration due to gravity (g') acting on the pendulum increases. The effective gravity can be calculated as:
\[ g' = g + a \]
where:
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( a \) is the upward acceleration of the elevator (given as \( 1 \, \text{m/s}^2 \)).

### Step 2: Calculate the Effective Gravity
Substituting the values:
\[ g' = 10 \, \text{m/s}^2 + 1 \, \text{m/s}^2 = 11 \, \text{m/s}^2 \]

### Step 3: Use the Formula for the Period of a Pendulum
The period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g'}} \]
where:
- \( L \) is the length of the pendulum (given as \( 1 \, \text{m} \)),
- \( g' \) is the effective gravity we calculated in Step 2.

### Step 4: Substitute the Values into the Formula
Now substituting \( L = 1 \, \text{m} \) and \( g' = 11 \, \text{m/s}^2 \):
\[ T = 2\pi \sqrt{\frac{1}{11}} \]

### Step 5: Calculate the Period
Calculating \( T \):
\[ T = 2\pi \sqrt{\frac{1}{11}} \approx 2\pi \cdot 0.301 \approx 1.884 \, \text{s} \]

### Step 6: Calculate the Frequency
The frequency \( f \) is the reciprocal of the period:
\[ f = \frac{1}{T} \]
Substituting for \( T \):
\[ f = \frac{1}{2\pi \sqrt{\frac{1}{11}}} \]

### Step 7: Simplify the Frequency Calculation
Using the approximation \( \pi \approx 3.14 \):
\[ f \approx \frac{1}{1.884} \approx 0.53 \, \text{Hz} \]

### Final Answer
Thus, the frequency of the pendulum is approximately:
\[ f \approx 0.528 \, \text{Hz} \]

### Conclusion
The correct answer is \( \text{D} \).

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