A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is

Options

A. 2 Hz
B. `(1)/(4) Hz`
C. `(1)/(2sqrt(2))Hz`
D. `(1)/(2)` Hz

Answer

To solve the problem, we need to find the frequency of vibration of an 8 kg block attached to two identical springs in parallel, given that a 1 kg block attached to one spring vibrates at a frequency of 1 Hz.

### Step-by-Step Solution:

1. **Understand the relationship between frequency, spring constant, and mass**:
The frequency of a mass-spring system is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
where \( f \) is the frequency, \( k \) is the spring constant, and \( m \) is the mass.

2. **Calculate the spring constant \( k \) for the 1 kg block**:
Given that the frequency \( f_1 \) for the 1 kg block is 1 Hz, we can set up the equation:
\[ 1 = \frac{1}{2\pi} \sqrt{\frac{k}{1}} \]
Squaring both sides gives:
\[ 1 = \frac{1}{4\pi^2} k \]
Rearranging gives:
\[ k = 4\pi^2 \]

3. **Determine the equivalent spring constant for the 8 kg block**:
Since the two springs are identical and connected in parallel, the equivalent spring constant \( k_{eq} \) is:
\[ k_{eq} = k + k = 2k = 2(4\pi^2) = 8\pi^2 \]

4. **Calculate the frequency \( f_2 \) for the 8 kg block**:
Now, we can use the frequency formula for the 8 kg block:
\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{k_{eq}}{m}} = \frac{1}{2\pi} \sqrt{\frac{8\pi^2}{8}} \]
Simplifying this gives:
\[ f_2 = \frac{1}{2\pi} \sqrt{\pi^2} = \frac{1}{2\pi} \cdot \pi = \frac{1}{2} \]

5. **Final answer**:
Thus, the frequency of vibration of the 8 kg block is:
\[ f_2 = \frac{1}{2} \text{ Hz} \]