A mass of 1.53 kg is attached to a spring and the system is undergoing simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm What is the total mechanical energy of the system? Select one: 0.955 0J 0.646 J 0.633 j 0.844

To find the total mechanical energy of the system, we can use the formula for the total mechanical energy in simple harmonic motion:<br /><br />\[ E = \frac{1}{2} k A^2 \]<br /><br />where:<br />- \( E \) is the total mechanical energy,<br />- \( k \) is the spring constant,<br />- \( A \) is the amplitude.<br /><br />First, we need to find the spring constant \( k \). We can use the formula for the angular frequency \( \omega \) in simple harmonic motion:<br /><br />\[ \omega = 2\pi f \]<br /><br />where:<br />- \( \omega \) is the angular frequency,<br />- \( f \) is the frequency.<br /><br />Given that the frequency \( f \) is 1.95 Hz, we can calculate the angular frequency \( \omega \):<br /><br />\[ \omega = 2\pi \times 1.95 \approx 12.27 \, \text{rad/s} \]<br /><br />Next, we can use the formula for the spring constant \( k \):<br /><br />\[ k = m \omega^2 \]<br /><br />where:<br />- \( m \) is the mass.<br /><br />Given that the mass \( m \) is 1.53 kg, we can calculate the spring constant \( k \):<br /><br />\[ k = 1.53 \times (12.27)^2 \approx 183.6 \, \text{N/m} \]<br /><br />Now, we can calculate the total mechanical energy \( E \) using the amplitude \( A \) of 7.50 cm (or 0.075 m):<br /><br />\[ E = \frac{1}{2} \times 183.6 \times (0.075)^2 \approx 0.646 \, \text{J} \]<br /><br />Therefore, the correct answer is:<br /><br />\[ \boxed{0.646 \, \text{J}} \]