Given:-
$t_0 = 2mm$ (Thickness of plate)
$Y = 8 × 10^{10} N/m^2$
$ρ = 2650 kg/m^3$
$f = 3 MHz = 3 ×10^6 MHz$
To find: -
Original Natural Frequency $(f_0)$
Change in thickness (∆t) for f = 3 MHz
Solution:-
- Original Natural Frequency:-
$$f_0 = \frac{1}{2t_0}\sqrt{\frac{Y}{ρ}}$$
$$ ∴ f_0 = \frac{1}{2 × 2 × 10^{-3}}\sqrt{\frac{8 × 10^{10}}{2650}}$$
$$ ∴ f_0 = 1.374 × 10^6 Hz$$
$$ i.e. f_0 = 1.374 MHz$$ ...Ans
- Change in thickness required:-
Let "t" be the thickness required
$$ f = \frac{1}{2t}\sqrt{\frac{Y}{ρ}}$$
$$ ∴ 3 × 10^6 = \frac{1}{2 × t}\sqrt{\frac{8 × 10^{10}}{2650}}$$
$$ ∴ t = 9.157 × 10^{-4} m = 0.9157 mm$$
$$ ∴ ∆t = t_0 - t = 2 - 0.9157$$
$$ ∴ ∆t = 1.0843 mm$$