Question

A quartz crystal of thickness 0.5 cm and length 5 cm is vibrating at resonance for its fundamental mode. Calculate the frequency of vibration of the crystal. Given: Young’s Modulus Y = 8x10¹⁰N/m², density of quartz ρ= 2.654x10³ kg/m³.

Answer 1

$$\begin{gathered} \mathrm{the}\mathrm{frequency}\mathrm{of}\mathrm{quartz}\mathrm{crystal}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \mathrm{\nu }=\frac{1}{2\mathrm{d}}\sqrt{\frac{\mathrm{Y}}{\mathrm{\rho }}},\mathrm{where}\mathrm{Y},\mathrm{d}\mathrm{and}\mathrm{\rho }\mathrm{are}\mathrm{Young}\text{'}\mathrm{s}\mathrm{modulas}, \\ \mathrm{thickness}\mathrm{of}\mathrm{quartz}\mathrm{crystal}\mathrm{and}\mathrm{density}\mathrm{of}\mathrm{crystal}\mathrm{respectively}. \\ \mathrm{Here}\mathrm{\rho }=\mathrm{density}=2.654\times {10}^3\mathrm{Kg}/{\mathrm{m}}^3,\mathrm{Y}=8\times {10}^{10}\mathrm{N}/{\mathrm{m}}^2\mathrm{and}\mathrm{d}=0.5\times {10}^{-2}\mathrm{m}. \\ \mathrm{Therefore}\mathrm{\nu }=\frac{1}{2\times 0.5\times {10}^{-2}}\sqrt{\frac{8\times {10}^{10}}{2.654\times {10}^3}} \\ ={10}^2\times \sqrt{30.14}\times {10}^3=5.49\times {10}^5\mathrm{Hz}=0.549\mathrm{MHz}. \\ \mathrm{Therefore}\mathrm{the}\mathrm{frequency}\mathrm{of}\mathrm{vibration}\mathrm{is}5.49\times {10}^5\mathrm{Hz}., \end{gathered}$$