**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

$$ f = \frac{1}{2t}\sqrt{\frac{Y}{ρ}}$$

$$ ∴ 3 × 10^6 = \frac{1}{2 × t}\sqrt{\frac{8 × 10^{10}}{2650}}$$

$$ ∴ t = 9.157 × 10^{-4} m = 0.9157 mm$$

$$ ∴ ∆t = t_0 - t = 2 - 0.9157$$

$$ ∴ ∆t = 1.0843 mm$$