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Find the natural frequency of vibration of quartz plate of thickness 2mm

Given Young's modulus of quartz $Y = 8 × 10^{10} N/m^2$, density is 2650 $kg/m^3$. Calculate the change in thickness required if the same plate is used to produce ultrasonic waves of frequency 3MHz.

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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:-

  1. 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

  1. Change in thickness required:-

Let "t" be the thickness required

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

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

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

$$ ∴ ∆t = t_0 - t = 2 - 0.9157$$

$$ ∴ ∆t = 1.0843 mm$$

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