Understanding Engine Resonance Speed Calculation

This question involves calculating the natural frequency of an automotive engine system, which is the speed at which resonance is likely to occur. Resonance happens when the frequency of an external force matches the system's natural frequency, leading to potentially large oscillations.

Calculating Total Stiffness

The engine is supported by 4 springs. The total stiffness ($k_{eq}$) is the sum of the stiffnesses of all springs because they are arranged in parallel, supporting the engine's mass.

• Stiffness of each front spring ($k_f$): $3 \text{ MN/m} = 3 \times 10^6 \text{ N/m}$
• Number of front springs: 2
• Stiffness of each rear spring ($k_r$): $4.5 \text{ MN/m} = 4.5 \times 10^6 \text{ N/m}
• Number of rear springs: 2

Total stiffness ($k_{eq}$) is calculated as:

$k_{eq} = (2 \times k_f) + (2 \times k_r)$
$k_{eq} = (2 \times 3 \times 10^6 \text{ N/m}) + (2 \times 4.5 \times 10^6 \text{ N/m})$
$k_{eq} = (6 \times 10^6 \text{ N/m}) + (9 \times 10^6 \text{ N/m})$
$k_{eq} = 15 \times 10^6 \text{ N/m}$

Calculating Natural Circular Frequency

The natural circular frequency ($\omega_n$) of a system is determined by its stiffness and mass. The formula is:

$\omega_n = \sqrt{\frac{k_{eq}}{m}}$

Given:

• Total stiffness ($k_{eq}$) = $15 \times 10^6 \text{ N/m}$
• Mass of the engine ($m$) = $135 \text{ kg}$

Substituting the values:

$\omega_n = \sqrt{\frac{15 \times 10^6 \text{ N/m}}{135 \text{ kg}}}$
$\omega_n = \sqrt{\frac{150 \times 10^5}{135}}$
$\omega_n = \sqrt{\frac{10 \times 10^5}{9}}$
$\omega_n = \sqrt{\frac{10^6}{9}}$
$\omega_n = \frac{1000}{3} \text{ rad/s}$

Converting Frequency to Revolutions Per Minute (RPM)

Resonance speed is usually expressed in RPM. First, we convert the natural circular frequency ($\omega_n$) to frequency ($f_n$) in Hertz (Hz) using the formula $\omega_n = 2 \pi f_n$.

$f_n = \frac{\omega_n}{2 \pi}$
$f_n = \frac{1000/3}{2 \pi}$
$f_n = \frac{1000}{6 \pi} \text{ Hz}$

To convert frequency from Hz to RPM, we multiply by 60 (since there are 60 seconds in a minute):

Speed (RPM) = $f_n \times 60$
Speed (RPM) = $\frac{1000}{6 \pi} \times 60$
Speed (RPM) = $\frac{1000 \times 10}{\pi}$
Speed (RPM) = $\frac{10000}{\pi}$

Therefore, the engine speed at which resonance is likely to occur is $\frac{10000}{\pi}$ RPM.