Being a punctual man, the lift operator of a skyscraper hung an exact pendulum clock on the lift wall to know the end of the working day. The lift moves with an upward and downward accelerations during the same time (according to a stationary clock), the magnitudes of the accelerations remaining unchanged.
Will the operator finish his working day in time, or will he work more (less) than required?

Explanation

To solve the problem, we need to analyze how the pendulum clock behaves when the lift is accelerating upwards and downwards.

### Step-by-step Solution:

1. **Understanding the Pendulum Clock**:
The time 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 and \( g \) is the acceleration due to gravity.

2. **Effect of Upward Acceleration**:
When the lift accelerates upwards with acceleration \( a \), the effective acceleration acting on the pendulum becomes \( g + a \). Therefore, the time period \( T_{\text{up}} \) for the pendulum clock in the upward moving lift is:
\[ T_{\text{up}} = 2\pi \sqrt{\frac{L}{g + a}} \]

3. **Effect of Downward Acceleration**:
Conversely, when the lift accelerates downwards with the same magnitude of acceleration \( a \), the effective acceleration acting on the pendulum becomes \( g - a \). Thus, the time period \( T_{\text{down}} \) for the pendulum clock in the downward moving lift is:
\[ T_{\text{down}} = 2\pi \sqrt{\frac{L}{g - a}} \]

4. **Time Measurement During Upward and Downward Movement**:
Let the lift move upwards for time \( t \) and downwards for the same time \( t \). The total time measured by the pendulum clock during these movements will be:
\[ n_1 = \frac{t}{T_{\text{up}}} = \frac{t}{2\pi \sqrt{\frac{L}{g + a}}} \]
\[ n_2 = \frac{t}{T_{\text{down}}} = \frac{t}{2\pi \sqrt{\frac{L}{g - a}}} \]

5. **Total Time Measured by the Pendulum Clock**:
The total time \( n \) measured by the pendulum clock during the entire cycle (up and down) is:
\[ n = n_1 + n_2 = \frac{t}{2\pi \sqrt{\frac{L}{g + a}}} + \frac{t}{2\pi \sqrt{\frac{L}{g - a}}} \]

6. **Comparison with Actual Time**:
The actual time elapsed according to a stationary clock during the same period is \( 2t \). We need to compare \( n \) with \( 2t \) to determine if the operator finishes his work on time, or works more or less.

7. **Simplifying the Comparison**:
To simplify the comparison, we can express \( n \) in terms of \( t \):
\[ n = t \left( \frac{1}{2\pi \sqrt{\frac{L}{g + a}}} + \frac{1}{2\pi \sqrt{\frac{L}{g - a}}} \right) \]
Since \( \sqrt{\frac{L}{g + a}} < \sqrt{\frac{L}{g}} < \sqrt{\frac{L}{g - a}} \), it follows that:
\[ \frac{1}{\sqrt{g + a}} > \frac{1}{\sqrt{g}} > \frac{1}{\sqrt{g - a}} \]
Hence, \( n < 2t \).

8. **Conclusion**:
Since \( n < 2t \), the operator will finish his working day in less time than required. Therefore, he will work more than the required time according to the stationary clock.