A metre stick swinging about its one end oscillates with frequency `f_(0)`. If the bottom half of the stick was cut off, then its new oscillation frequency will be :

Options

A. `f_(0)`
B. `sqrt(2)f_(0)`
C. `2f_(0)`
D. `2sqrt(2)f_(0)`

Explanation

To solve the problem of finding the new oscillation frequency of a metre stick after cutting off its bottom half, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Setup**:
- We have a metre stick of length \( L \) (1 metre) swinging about one end.
- The initial frequency of oscillation is given as \( f_0 \).

2. **Frequency Formula**:
- The frequency of oscillation for a physical pendulum is given by:
\[ f = \frac{1}{2\pi} \sqrt{\frac{mgL}{I}} \]
- Here, \( m \) is the mass, \( g \) is the acceleration due to gravity, \( L \) is the distance from the pivot to the center of mass, and \( I \) is the moment of inertia.

3. **Determine the Initial Parameters**:
- For the full metre stick, the length \( L = 1 \) m.
- The center of mass of a uniform stick is at \( \frac{L}{2} = 0.5 \) m from the pivot.
- The moment of inertia \( I \) for a stick about one end is given by:
\[ I = \frac{1}{3} m L^2 \]

4. **Calculate the Initial Frequency \( f_0 \)**:
- Substitute \( L = 1 \) m into the frequency formula:
\[ f_0 = \frac{1}{2\pi} \sqrt{\frac{mg \cdot 0.5}{\frac{1}{3} m (1)^2}} \]
- Simplifying this gives:
\[ f_0 = \frac{1}{2\pi} \sqrt{\frac{mg \cdot 0.5}{\frac{1}{3} m}} = \frac{1}{2\pi} \sqrt{\frac{3g}{2}} \]

5. **Determine the New Parameters After Cutting the Stick**:
- When the bottom half is cut off, the new length \( L' = \frac{1}{2} \) m.
- The new center of mass is now at \( \frac{L'}{4} = \frac{1/2}{4} = \frac{1}{8} \) m from the pivot.
- The new mass is \( m' = \frac{m}{2} \).

6. **Calculate the New Moment of Inertia \( I' \)**:
- The moment of inertia for the new stick is:
\[ I' = \frac{1}{3} m' (L')^2 = \frac{1}{3} \left(\frac{m}{2}\right) \left(\frac{1}{2}\right)^2 = \frac{1}{3} \cdot \frac{m}{2} \cdot \frac{1}{4} = \frac{m}{24} \]

7. **Calculate the New Frequency \( f' \)**:
- Substitute the new values into the frequency formula:
\[ f' = \frac{1}{2\pi} \sqrt{\frac{m' g \cdot \frac{L'}{4}}{I'}} = \frac{1}{2\pi} \sqrt{\frac{\frac{m}{2} g \cdot \frac{1}{2}}{\frac{m}{24}}} \]
- Simplifying gives:
\[ f' = \frac{1}{2\pi} \sqrt{\frac{\frac{mg}{4}}{\frac{m}{24}}} = \frac{1}{2\pi} \sqrt{6g} \]

8. **Relate the New Frequency to the Old Frequency**:
- We can relate \( f' \) to \( f_0 \):
\[ \frac{f'}{f_0} = \sqrt{\frac{6g}{\frac{3g}{2}}} = \sqrt{4} = 2 \]
- Therefore, \( f' = 2 f_0 \).

### Final Answer:
The new oscillation frequency after cutting the bottom half of the stick will be:
\[ f' = 2 f_0 \]