To determine the maximum amplitude [tex]\( A \)[/tex] that allows a particle to remain on a plank undergoing simple harmonic motion (SHM) with a given frequency, we can use the relationship between acceleration in SHM and gravitational acceleration.
The particle will start to leave the plank when the maximum acceleration due to SHM equals the acceleration due to gravity. The maximum acceleration [tex]\( a_{\text{max}} \)[/tex] in SHM is given by the formula:
[tex]\[ a_{\text{max}} = A \omega^2 \][/tex]
where [tex]\( A \)[/tex] is the amplitude and [tex]\( \omega \)[/tex] is the angular frequency.
1. Identify the given data:
- Frequency of SHM, [tex]\( f = \frac{3}{\pi} \)[/tex] Hz
- Acceleration due to gravity, [tex]\( g = 9.81 \)[/tex] m/s[tex]\(^2\)[/tex]
2. Determine the angular frequency [tex]\(\omega\)[/tex]:
[tex]\[ \omega = 2 \pi f \][/tex]
3. Substitute the given frequency into the formula:
[tex]\[ \omega = 2 \pi \left(\frac{3}{\pi}\right) \][/tex]
[tex]\[ \omega = 2 \pi \cdot \frac{3}{\pi} \][/tex]
[tex]\[ \omega = 2 \cdot 3 \][/tex]
[tex]\[ \omega = 6 \text{ rad/s} \][/tex]
4. For the particle to remain on the plank, set the maximum SHM acceleration equal to the gravitational acceleration:
[tex]\[ A \omega^2 = g \][/tex]
5. Solve for the amplitude [tex]\( A \)[/tex]:
[tex]\[ A = \frac{g}{\omega^2} \][/tex]
6. Substitute the known values for [tex]\( g \)[/tex] and [tex]\( \omega \)[/tex]:
[tex]\[ A = \frac{9.81}{6^2} \][/tex]
[tex]\[ A = \frac{9.81}{36} \][/tex]
[tex]\[ A \approx 0.2725 \text{ m} \][/tex]
Thus, the maximum amplitude [tex]\( A \)[/tex] for the particle to not leave the plank is approximately [tex]\( 0.2725 \)[/tex] meters.
Comparing this value with the given choices,
- [tex]\( \frac{2}{9} \)[/tex] m converts to approximately [tex]\( 0.222 \)[/tex] m, closely matching the calculated value.
Therefore, the correct answer is:
C. [tex]\(\frac{2}{9} \text{ m}\)[/tex]
