A weightless spring has a force constant k oscillates with frequency f when a mass m is suspended from it The spring is cut into three equal parts and a mass 3 m is suspended from it The frequency of oscillation of one part will now become

Options

A. f
B. 2 f
C. f / 3
D. 3 f

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the initial conditions
The initial frequency \( f \) of the spring with a mass \( m \) is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
where \( k \) is the spring constant.

**Hint:** Remember that the frequency of oscillation depends on both the mass attached and the spring constant.

### Step 2: Analyze the effect of cutting the spring
When the spring is cut into three equal parts, the length of each part becomes \( \frac{L}{3} \) (where \( L \) is the original length of the spring). The spring constant \( k' \) of each part is related to the original spring constant \( k \) by the formula:
\[ k' = 3k \]
This is because the spring constant is inversely proportional to the length of the spring.

**Hint:** Cutting a spring into shorter lengths increases the spring constant.

### Step 3: Determine the new mass
Now, a mass of \( 3m \) is suspended from one of the cut parts of the spring.

**Hint:** Note how the mass has changed from \( m \) to \( 3m \).

### Step 4: Calculate the new frequency
The new frequency \( f' \) can be calculated using the new spring constant \( k' = 3k \) and the new mass \( m' = 3m \):
\[ f' = \frac{1}{2\pi} \sqrt{\frac{k'}{m'}} = \frac{1}{2\pi} \sqrt{\frac{3k}{3m}} = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = f \]

**Hint:** The new frequency simplifies to the original frequency because the factors of 3 cancel out.

### Conclusion
Thus, the frequency of oscillation of one part of the spring with mass \( 3m \) becomes:
\[ f' = f \]

**Final Answer:** The frequency remains the same, \( f \).