A spring mass system oscillates with a frequency v. If it is taken in an elavator slowly accelerating upward, the frequency will

Options

A. increases
B. decreases
C. remain same
D. become zero

Answer

To solve the problem, we need to analyze how the frequency of a spring-mass system is affected when it is placed in an elevator that is accelerating upward.

### Step-by-Step Solution:

1. **Understanding the Spring-Mass System**:
The frequency \( f \) of a spring-mass system is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
where \( k \) is the spring constant and \( m \) is the mass of the object attached to the spring.

2. **Effect of the Elevator's Acceleration**:
When the elevator accelerates upward, the effective gravitational force acting on the mass changes. However, the spring constant \( k \) and the mass \( m \) remain unchanged.

3. **Time Period Calculation**:
The time period \( T \) of the oscillation is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
The frequency is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]

4. **Independence from Acceleration**:
The key point here is that the frequency \( f \) does not depend on the gravitational acceleration \( g \) or any additional forces acting on the mass due to the elevator's motion. The formula for frequency remains the same regardless of the upward acceleration.

5. **Conclusion**:
Since the frequency is independent of the acceleration of the elevator, the frequency of the spring-mass system will remain the same even when the elevator is accelerating upward.

Thus, the correct answer is that the frequency remains the same.

### Final Answer:
The frequency will **remain the same**.

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