Question 14.3: A vibrating floor. A large motor in a factory causes the flo......
A vibrating floor. A large motor in a factory causes the floor to vibrate at a frequency of 10 Hz. The amplitude of the floor’s motion near the motor is about 3.0 mm. Estimate the maximum acceleration of the floor near the motor.
APPROACH Assuming the motion of the floor is roughly SHM we can make an estimate for the maximum acceleration using Eq. 14-9b.
a_{\max }=\omega^{2} A=\frac{k}{m} A (14-9b)
Given \omega=2 \pi f=(2 \pi)\left(10 \mathrm{~s}^{-1}\right)=62.8 \,\mathrm{rad} / \mathrm{s}, then Eq. 14-9b gives
a_{\max }=\omega^{2} A=(62.8 \,\mathrm{rad} / \mathrm{s})^{2}(0.0030 \mathrm{~m})=12 \mathrm{~m} / \mathrm{s}^{2} .
NOTE The maximum acceleration is a little over g, so when the floor accelerates down, objects sitting on the floor will actually lose contact momentarily, which will cause noise and serious wear.