If the frequency of an object in unifrom circular motion is doubled, its acceleration becomes

Options

A. Two times
B. Four times
C. Half
D. one fourth

Answer

To solve the problem, we need to understand the relationship between the frequency of an object in uniform circular motion and its acceleration.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
In uniform circular motion, the centripetal acceleration \( a \) is given by the formula:
\[ a = \frac{v^2}{r} \]
where \( v \) is the linear velocity and \( r \) is the radius of the circular path.

2. **Relating Velocity to Frequency**:
The linear velocity \( v \) can also be expressed in terms of the angular frequency \( \omega \) and the radius \( r \):
\[ v = r \omega \]
The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:
\[ \omega = 2\pi f \]

3. **Substituting Angular Frequency**:
Substituting \( \omega \) into the velocity equation:
\[ v = r (2\pi f) = 2\pi r f \]

4. **Substituting Velocity into Acceleration**:
Now substituting \( v \) back into the acceleration formula:
\[ a = \frac{(2\pi r f)^2}{r} \]
Simplifying this gives:
\[ a = \frac{4\pi^2 r f^2}{r} = 4\pi^2 f^2 \]

5. **Effect of Doubling Frequency**:
If the frequency \( f \) is doubled, then the new frequency \( f' \) is:
\[ f' = 2f \]
Now we can find the new acceleration \( a' \):
\[ a' = 4\pi^2 (f')^2 = 4\pi^2 (2f)^2 = 4\pi^2 \cdot 4f^2 = 16\pi^2 f^2 \]

6. **Comparing New Acceleration to Original Acceleration**:
The original acceleration \( a \) was:
\[ a = 4\pi^2 f^2 \]
Therefore, the new acceleration \( a' \) is:
\[ a' = 4 \times (4\pi^2 f^2) = 4a \]

### Conclusion:
Thus, if the frequency of an object in uniform circular motion is doubled, its acceleration becomes **four times** the original acceleration.

### Final Answer:
The correct option is **four times** (Option B).