To determine the number of complete oscillations made by the spring-block system in one minute, we first need to calculate the angular frequency of the oscillation. The system consists of two blocks connected by a spring, and we can treat it as a simple harmonic oscillator. Let's break down the steps to find the solution.
Understanding the System
In this scenario, we have two blocks with masses \( m_1 = 2 \, \text{kg} \) and \( m_2 = 3 \, \text{kg} \) connected by a spring with a force constant \( k = 300 \, \text{N/m} \). The total mass of the system can be calculated as:
- Total mass \( m = m_1 + m_2 = 2 \, \text{kg} + 3 \, \text{kg} = 5 \, \text{kg} \)
Calculating Angular Frequency
The angular frequency \( \omega \) of a mass-spring system is given by the formula:
\( \omega = \sqrt{\frac{k}{m}} \)
Substituting the values we have:
\( \omega = \sqrt{\frac{300 \, \text{N/m}}{5 \, \text{kg}}} \)
Calculating this gives:
\( \omega = \sqrt{60} \)
Given that \( \pi \approx \sqrt{10} \), we can approximate \( \sqrt{60} \) as follows:
\( \sqrt{60} = \sqrt{6 \times 10} = \sqrt{6} \cdot \sqrt{10} \approx 2.45 \cdot 3.16 \approx 7.76 \, \text{rad/s} \)
Finding the Frequency
The frequency \( f \) of the oscillation is related to the angular frequency by the formula:
\( f = \frac{\omega}{2\pi} \)
Substituting our value of \( \omega \):
\( f = \frac{7.76}{2\pi} \approx \frac{7.76}{2 \cdot 3.16} \approx \frac{7.76}{6.32} \approx 1.23 \, \text{Hz} \)
Calculating the Number of Oscillations
Now that we have the frequency, we can find the number of complete oscillations in one minute. Since there are 60 seconds in a minute, we multiply the frequency by 60:
\( \text{Number of oscillations} = f \times 60 \approx 1.23 \, \text{Hz} \times 60 \approx 73.8 \)
Since we are interested in complete oscillations, we round this down to the nearest whole number:
\( \text{Complete oscillations} \approx 73 \)
Final Result
Therefore, the spring-block system will complete approximately 73 complete oscillations in one minute.
