Answer
The fractional half-width is $~~0.040$
Work Step by Step
In part (a), we found that $~~\omega_d = 224~rad/s$ at resonance.
In part (c), we found that $~~\omega_{d1} = 219~rad/s$
In part (d), we found that $~~\omega_{d2} = 228~rad/s$
We can find the fractional half-width:
$\frac{\omega_{d2}-\omega_{d1}}{\omega_d} = \frac{228~rad-219~rad/s}{224~rad/s} = 0.040$
The fractional half-width is $~~0.040$
