Answer :
The damping constant of the oscillator, as the amplitude decreases to 44.2% of its initial value in 46.5 seconds, is approximately [tex]0.0261 s^{(-1)}[/tex].
The damping constant [tex](\(\gamma\))[/tex] can be found using the formula:
[tex]\[ \gamma = -\frac{1}{T} \ln\left(\frac{A_t}{A_0}\right) \][/tex]
where:
- [tex]\( \gamma \)[/tex] is the damping constant,
- [tex]\( T \)[/tex] is the time taken for the amplitude to decrease to a fraction of its initial amplitude,
- [tex]\( A_t \)[/tex] is the final amplitude,
- [tex]\( A_0 \)[/tex] is the initial amplitude.
In this case, [tex]\( T = 46.5 \) s, \( A_t = 0.442 \times A_0 \), and \( f_0 = 1.58 \) Hz[/tex].
First, calculate the final amplitude [tex]\( A_t \)[/tex]:
[tex]\[ A_t = 0.442 \times A_0 \][/tex]
Next, use the formula for the damping constant:
[tex]\[ \gamma = -\frac{1}{T} \ln\left(\frac{A_t}{A_0}\right) \][/tex]
Now, substitute the values and solve for [tex]\( \gamma \)[/tex]:
[tex]\[ \gamma = -\frac{1}{46.5} \ln\left(\frac{0.442 \times A_0}{A_0}\right) \][/tex]
[tex]\[ \gamma = -\frac{1}{46.5} \ln(0.442) \][/tex]
[tex]\[ \gamma \approx 0.0261 \, \text{s}^{-1} \][/tex]
Therefore, the damping constant of the oscillation is approximately [tex]\(0.0261 \, \text{s}^{-1}\)[/tex].