Question 10.5: A vibration test table is set up to produce a constant PSD l......
A vibration test table is set up to produce a constant PSD level of 0.02 g²/Hz over the frequency range 10–1000 Hz. Find
(a) The RMS acceleration level of the table.
(b) The RMS displacement of the table, and the total movement if the displacement is Gaussian and limited to ± 3 times the RMS value.
Let S_{\mathrm{g}} = acceleration PSD level in g²/Hz;
S_{\ddot{\mathrm{x}} } = acceleration PSD level in m/s²/Hz;
S_{\mathrm{x}} = displacement PSD level in m²/Hz;
\sigma _{\mathrm{g}} = RMS level in g units (1 g = 9.81 m/s²);
\sigma _{\mathrm{x}} = RMS level in meters.
Part (a):
From Eq. (10.30),
From Eq. (A),
\sigma_{\mathrm{g}}^{2} = \int_{10}^{1000}{0.02 df = \left[0.02f\right]_{10}^{1000} } = 0.02 (1000 – 10) = 19.8 \mathrm{g}^{2}.RMS acceleration of table = \sigma _{\mathrm{g}} = \sqrt{19.8} = 4.45 \mathrm{g}
Part (b):
In this case, since acceleration is to be converted to displacement, consistent units must be used, and the acceleration must be expressed in m/s². Since 1 g = 9.81 m/s²: S_{\ddot{\mathrm{x}} } = (9.81 ^{2} S_{\mathrm{g}}) = (9.81 ^{2} \times 0.02) = 1.925 m/s²/Hz , in the range 10–1000 Hz, and zero elsewhere. Then from Eq. (10.39b):
From Eq. (10.30) and Eq. (B),
\sigma_{\mathrm{x}}^{2} = \int_{f_{1}}^{f_{2}}{S_{\mathrm{x}} df} = (1.235 \times 10^{-3}) \int_{10}^{1000}{f^{-4}df} = (1.235 \times 10^{-3}) \left[-\frac{1}{4}f^{-3} \right] ^{1000}_{10} \\ =(0.3088 \times 10^{-6}) \mathrm{m² and } \sigma_{\mathrm{x}} = 0.556 \times 10^{-3} m = 0.556 mm.If the displacement has a Gaussian amplitude distribution limited to \pm 3 \sigma_{\mathrm{x}} the total movement is 6 × 0.556 mm = 3.34 mm.