Solveeit Logo

Question

Question: A piezoelectric quartz crystal of thickness \(0.005m\) is vibrating in resonance conditions. Calcula...

A piezoelectric quartz crystal of thickness 0.005m0.005m is vibrating in resonance conditions. Calculate the fundamental frequency f0{f_0} for quartz (Y=8×1010Nm2Y = 8 \times {10^{10}}N{m^{ - 2}} and ρ=2.56×103\rho = 2.56 \times {10^3} )
A) 5.5MHz5.5MHz
B) 55MHz55MHz
C) 0.55MHz0.55MHz
D) 5.5kHz5.5kHz

Explanation

Solution

First we have to find the speed of sound waves in quartz crystals by using the relation between frequency and velocity of waves. We know that the fundamental frequency length or thickness of the medium must be half of the wavelength of the wave.

Step by step solution:
Step 1
Velocity of wave in quartz crystal v=Yρv = \sqrt {\dfrac{Y}{\rho }}
Where YY \Rightarrow young modulus for medium
ρ\rho \Rightarrow Density of medium
v=8×10102.65×103\Rightarrow v = \sqrt {\dfrac{{8 \times {{10}^{10}}}}{{2.65 \times {{10}^3}}}}
Solving this
v=3.01×107\Rightarrow v = \sqrt {3.01 \times {{10}^7}} M/sec
Step 2
We know the relation between velocity of wave, frequency and wavelength is given as
v=fλv = f\lambda .................. (1)
Where ff \Rightarrow frequency of wave
λ\lambda \Rightarrow Wavelength of wave
For fundamental frequency the length or thickness must be equal to half of wavelength
L=λ2\Rightarrow L = \dfrac{\lambda }{2}
λ=2L\Rightarrow \lambda = 2L
Where LL \Rightarrow thickness of quartz crystal
From equation (1)
f0=vλ\Rightarrow {f_0} = \dfrac{v}{\lambda }
f0=v2L\Rightarrow {f_0} = \dfrac{v}{{2L}}
Put the value of velocity and thickness L=0.005mL = 0.005m
f0=12×0.0053.01×107\Rightarrow {f_0} = \dfrac{1}{{2 \times 0.005}}\sqrt {3.01 \times {{10}^7}}
f0=5.4×10310×103\Rightarrow {f_0} = \dfrac{{5.4 \times {{10}^3}}}{{10 \times {{10}^{ - 3}}}}
Further solving
f0=0.54×106\Rightarrow {f_0} = 0.54 \times {10^6}
Hence
f0=0.55MHz\therefore {f_0} = 0.55MHz

Hence option C is correct.

Note: By this simple method we can calculate the fundamental frequency .We use here for fundamental frequency the wavelength must be double of length of medium as shown in figure

For fundamental vibration L=λ2L = \dfrac{\lambda }{2} it is clear from the above diagram.