To find the mass of the object attached to the spring, we follow these steps:
### Step 1: Calculate the Period of Oscillations
The period [tex]\( T \)[/tex] is the time taken for one complete oscillation. Given:
- The total number of oscillations [tex]\( N \)[/tex] = 15.0
- The total time [tex]\( t \)[/tex] = 9.24 seconds
The period [tex]\( T \)[/tex] is given by:
[tex]\[ T = \frac{t}{N} \][/tex]
Substituting the given values:
[tex]\[ T = \frac{9.24 \, \text{s}}{15.0} = 0.616 \, \text{s} \][/tex]
### Step 2: Find the Angular Frequency [tex]\( \omega \)[/tex]
The angular frequency [tex]\( \omega \)[/tex] is related to the period [tex]\( T \)[/tex] by the formula:
[tex]\[ \omega = \frac{2 \pi}{T} \][/tex]
Using the calculated period [tex]\( T = 0.616 \, \text{s} \)[/tex]:
[tex]\[ \omega = \frac{2 \pi}{0.616 \, \text{s}} = 10.19997614801881 \, \text{rad/s} \][/tex]
### Step 3: Use the Angular Frequency to Find the Mass
The relationship between the angular frequency [tex]\( \omega \)[/tex], the spring constant [tex]\( k \)[/tex], and the mass [tex]\( m \)[/tex] is given by:
[tex]\[ \omega = \sqrt{\frac{k}{m}} \][/tex]
Rearranging this equation to solve for [tex]\( m \)[/tex]:
[tex]\[ m = \frac{k}{\omega^2} \][/tex]
Given the spring constant [tex]\( k = 88.7 \, \text{N/m} \)[/tex] and the calculated angular frequency [tex]\( \omega = 10.19997614801881 \, \text{rad/s} \)[/tex]:
[tex]\[ m = \frac{88.7 \, \text{N/m}}{(10.19997614801881 \, \text{rad/s})^2} \][/tex]
[tex]\[ m = 0.8525606962596449 \, \text{kg} \][/tex]
### Conclusion
The mass of the object is:
[tex]\[ m \approx 0.853 \, \text{kg} \][/tex]
Thus, the object's mass is approximately 0.853 kg.
