A force of 6.4 N stretches a vertical spring by 0.1 m. The mass (in kg) that must be suspended from the spring so that it oscillates with a time period of `(pi)/(4)` second.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the Spring Constant (k)
We know that the force (F) applied to stretch the spring is given by Hooke's Law:
\[ F = k \cdot x \]
where:
- \( F = 6.4 \, \text{N} \) (force applied)
- \( x = 0.1 \, \text{m} \) (stretch in the spring)

Rearranging the formula to find the spring constant \( k \):
\[ k = \frac{F}{x} \]

Substituting the values:
\[ k = \frac{6.4 \, \text{N}}{0.1 \, \text{m}} = 64 \, \text{N/m} \]

### Step 2: Use the Time Period Formula
The time period \( T \) for a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

We are given that the time period \( T = \frac{\pi}{4} \, \text{s} \). We can set up the equation:
\[ \frac{\pi}{4} = 2\pi \sqrt{\frac{m}{k}} \]

### Step 3: Simplify the Equation
Dividing both sides by \( \pi \):
\[ \frac{1}{4} = 2 \sqrt{\frac{m}{k}} \]

Now, divide both sides by 2:
\[ \frac{1}{8} = \sqrt{\frac{m}{k}} \]

### Step 4: Square Both Sides
Squaring both sides to eliminate the square root:
\[ \left(\frac{1}{8}\right)^2 = \frac{m}{k} \]
\[ \frac{1}{64} = \frac{m}{k} \]

### Step 5: Solve for Mass (m)
Now, rearranging to find \( m \):
\[ m = \frac{k}{64} \]

Substituting the value of \( k \) we found earlier:
\[ m = \frac{64 \, \text{N/m}}{64} = 1 \, \text{kg} \]

### Final Answer
The mass that must be suspended from the spring is \( 1 \, \text{kg} \).

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