Let's break down the steps to determine the frequency of oscillation for the given mass-spring system.
### Step-by-Step Solution
1. Given Data:
- Mass, [tex]\( m = 8.0 \)[/tex] kg
- Stretch (displacement), [tex]\( x = 2.2 \)[/tex] cm, which needs to be converted into meters:
[tex]\[
x = 2.2 \text{ cm} = 2.2 \times 10^{-2} \text{ m} = 0.022 \text{ m}
\][/tex]
- Acceleration due to gravity, [tex]\( g = 9.81 \)[/tex] m/s[tex]\(^2\)[/tex]
2. Find the Spring Constant (k):
Using Hooke's Law, where the force [tex]\( F \)[/tex] exerted by the mass is balanced by the restoring force of the spring:
[tex]\[
F = kx \quad \text{and} \quad F = mg
\][/tex]
Therefore,
[tex]\[
mg = kx
\][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[
k = \frac{mg}{x}
\][/tex]
Substituting the given values:
[tex]\[
k = \frac{8.0 \text{ kg} \times 9.81 \text{ m/s}^2}{0.022 \text{ m}}
\][/tex]
[tex]\[
k = 3567.27 \text{ N/m}
\][/tex]
3. Determine the Angular Frequency [tex]\( \omega \)[/tex]:
The angular frequency [tex]\( \omega \)[/tex] is given by:
[tex]\[
\omega = \sqrt{\frac{k}{m}}
\][/tex]
[tex]\[
\omega = \sqrt{\frac{3567.27 \text{ N/m}}{8.0 \text{ kg}}}
\][/tex]
[tex]\[
\omega = 21.12 \text{ rad/s}
\][/tex]
4. Convert Angular Frequency to Frequency:
Frequency [tex]\( f \)[/tex] in Hz is given by:
[tex]\[
f = \frac{\omega}{2\pi}
\][/tex]
[tex]\[
f = \frac{21.12 \text{ rad/s}}{2\pi}
\][/tex]
[tex]\[
f = 3.361 \text{ Hz}
\][/tex]
### Conclusion
Hence, the frequency of oscillation for the given mass-spring system is approximately [tex]\( 3.4 \)[/tex] Hz.
So, the correct answer from the given options is [tex]\( 3.4 \)[/tex] Hz.
