Ultrasonics: Important Solved Problems

ANNA UNIVERSITY SOLVED PROBLEMS

 

 

1. Calculate the frequency to which a piezo electric oscillator circuit should be tuned so that a Piezo‒electric crystal of thickness 0.1cm vibrates in its fundamental mode to general ultrasonic waves. (Young's Modulus and density of material of crystal are 80 Gpa and 2654 kgm^‒3?

Solution

The frequency of vibration is given by f = P/2t . √(E/ρ)

For fundamental mode P = 1.

f = 1/2t . √(E/ρ)

f = 1/(2×0.1×10⁻²) . √( 80×10⁹ / 2654 )

f = 2.7451×10⁶ Hz

The frequency of the oscillator circuit = 2.7451×10⁶ Hz.

 

2. A quartz crystal of thickness 0.001 metre vibrates in its fundamental frequency. Calculate its frequency. Given that E=7.9×10¹⁰ N/m² and ρ =2650 kg/m³ for quartz.

Given data

 t = 1×10^‒3 m,

E=7.9×10¹⁰ N/m²;

ρ=2650 kg/m³

Solution

Formula

The frequency of vibration f = P/2t . √(E/ρ)

For fundamental mode of frequency P = 1

f= 1/(2×1×10^‒3) . √[7.9×10¹⁰ /2650]

f= 500 × 29.811 × 10⁶

f= 500 × 5459.945

f= 2.7299 × 10⁶ Hz

Frequency of vibrating crystal = 2.729 MHz

 

3. A quartz crystal in an ultrasonic interferometer produces stationary waves of frequency 1.5 MHz. If the distance between 6 consecutive nodes is 2.75 mm, find the velocity of the ultrasonic wave.

Solution

Given data

(i) Frequency of ultrasonics (v_u) = 1.5 MHz. = 1.5 × 10⁶ Hz.

 (ii) Distance between 6 consecutive nodes = 2.75 mm

i.e., 5d = 2.75 mm

d = (2.75/5) × 10^‒3 m

d = 5.5 x 10^‒4 m

From the above figure, we can write wavelength λ_u = 2d

 λ_u= 2 × 5.5 × 10^‒4 m

(or)

λ_u=  1.1 × 10^‒3 m

Velocity of ultrasonics = Frequency × Wavelength

v = v_uλ_u

v = 1.5 × 10⁶ × 1.1 × 10^‒3 m

v = 1650 m/sec.

Velocity of ultrasonics (v) = 1650 m/sec.

 

4. A ultrasonic generator consists of a quartz plate of thickness 0.7 mm and density 2800 kg/m³. Find the fundamental frequency of ultrasonic waves if the Young's modulus of quartz is 8.8 × 10¹⁰ N/m².

Given Data:

l = 0.7 mm = 0.7 × 10^‒3m

 ρ = 2800 kg/m³

P = 1

E = 8.8 × 10¹⁰ Nm^‒2

Formula:

Fundamental frequency (f) = P/2l . √(E/ρ)

 f = 1/(2×0.7×10^‒3) . √(8.8×10¹⁰ / 2800)

 f = 4.004 × 10⁶ Hz

(or) f=4.004 MHz

The fundamental frequency of Ultrasonic waves = 4.004 MHz

 

5. A quartz crystal of thickness 0.001 m is vibrating at resonance. Calculate the fundamental frequency. Density of quartz = 2.650 × 10³ kg/m³ and Young's modulus for quartz = 7.9 × 10¹⁰ Nm^‒2.

Given Data:

l=0.001m

p=2.650×10³ kg/m³

E=7.9×10¹⁰ Nm^‒2

P=1

Formula:

Fundamental frequency (f) =  P/2l . √(E/ρ)

= 1/(2 × 0.001) .  √ [7.9×10¹⁰ / 2.650×10³]

 f=2.7299×10⁶ Hz

The fundamental frequency = 2.7299 MHz

 

6. Longitudinal standing waves are set up in a quartz plate with antinodes at opposite faces. The fundamental frequency of vibration is given by the relation f = 2.87×10³/t where f is in Hz and t is in metre.

Compute (i) Young's modulus of quartz plate

(ii) The thickness of the plate required for a frequency of 1300 KHz. The density of quartz is 2660 kg m^‒3.

Solution

(i) The frequency of vibration is given by f =  P/2t . √(E/ρ)

For fundamental frequency P = 1

f =  1/2t . √(E/ρ)

Substituting the value of 'f', we have

2.87×10³ / t =  1/2t . √(E/2660)

E = 4 × 2660 × (2.87×10³)²

= 8.76406×10¹⁰ Nm^‒2

The Young modulus of the quartz plate = 8.76406×10¹⁰ Nm^‒2

(ii) The frequency of vibration = 1300 KHz

we know that f =  1/2t . √(E/ρ)

 (or)

Thickness (t) = 1/2f . √(E/ρ)

= 1 / (2x1300x10³) . √(8.76406×10¹⁰ / 2660)

= 2.2076 × 10^‒3 m

The thickness of the crystal = 2.2076×10^‒3 m.

 

7. An ultrasound pulse sent by a source in sea is reflected by a submerged target at a distance 597.50m and reaches the source after 0.83 seconds. Find the velocity of sound in sea water.

Given Data:

d = 597.50m

t = 0.83 sec

Formula:

Velocity (v) = 2d / t

Velocity (v) = 2×597.5 / 0.83

Velocity (v) = 1439.75 m/s

Velocity of sound in sea water = 1439.75 m/s

 

8. A quartz crystal with a thickness of 0.5 mm and a density of 2650 kg.m^‒3 vibrates longitudinally producing ultrasonic waves. Find the fundamental frequency of vibration if the Young's modulus of quartz is 7.9×10¹⁰ Nm^‒2.

Given data:

t = 0.5 × 10^‒3 m

ρ = 2650 kgm^‒3

E = 7.9 × 10¹⁰ Nm^‒2

Formulae:

f =  P/2t . √(E/ρ)

for fundamental mode P=1

f =  1/2t . √(E/ρ)

(or) f =  1/(2 × 0.5 × 10^‒3) . √(7.9 × 10¹⁰ / 2650)

f =  [1× (5459.97) ] / [ 1× 10^‒3 ]

f = 5.45997 × 10⁶ Hz

 The fundamental frequency= 5.45997 MHz

 

9. Calculate the fundamental frequency of a quartz crystal of thickness 1.5mm which is vibrating at resonance. Given Young's modulus for quartz 7.9×10¹⁰Nm^‒2 and density of quartz = 2650 kg m^‒3.

Solution

The frequency of vibration f =  P/2t . √(E/ρ)

Where P=1,2,3 etc for fundamental, first overtone, second overtone respectively,

Here P= 1

f = 1/[2×1.5×10^‒3]  .  √(7.9×10¹⁰ / 2650)

f = 1.8199 × 10⁶ Hz

The fundamental frequency of the quartz crystal = 1.8199×10⁶ Hz or 1.8199 MHz

 

ADDITIONAL SOLVED PROBLEMS

 

1. The velocity of ultrasonics in steel is 5000m/s. An ultrasonic beam is used to determine the thickness of a steel plate. It is noticed that the difference between two adjacent harmonic frequencies is 60 KHz. Determine the thickness of the steel plate.

Solution

The frequency of ultrasonics f = v / 2d

Let f_n and f_n‒1 be the two adjacent harmonic frequencies.

f_n ‒ f_n‒1 = v / 2d

d = v / 2(f_n ‒ f_n‒1)

The difference between the two harmonic frequency = f_n ‒ f_n‒1

= 60 KHz

= 60 × 10³ Hz

d = 5000 / 2×60×10³

 = 0.04166 m

The thickness of the steel plate is 0.04166 m.

 

2. Find the depth of a submerged submarine if an ultrasonic wave is received after 0.33 sec. from the time of transmission.

Given:

The velocity of ultrasonic waves in sea water = 1440 m/s.

Solution

Velocity of ultrasonics in sea water, v = 1440 m/s.

Time taken between transmitted and received ultrasonic waves t=0.33 sec.

We know velocity (v) = Distance (d) / Time (t)

Total distance travelled by ultrasonics = v × t

Distance (d) = 1440 × 0.33

Distance (d) = 475.2 m

Here, the total distance travelled by ultrasonics waves refers to the distance travelled (i) From source to the submarine and (ii) From the submarine to the source i.e., equal to twice the depth of the submarine from the transmitting end.

Depth of submerged submarine = 475.2 / 2

= 237.6 metres

Depth of the submerged submarine = 237.6 metres.

 

3. An ultrasonic interferometer used to measure the velocity in sea water. If the distance between two constructive antinodes is 0.55mm. Compute the velocity of the waves in the sea water. The frequency of the crystal is 1.5 MHz.

Solution

Distance between two antinodes = λ/2

= 0.55 mm.

λ/2 = 0.55 × 10^‒3 m

λ = 1.1 × 10^‒3 m

Velocity of ultrasonics = Frequency × wavelength

= 1.5 × 10⁶ × 1.1 × 10^‒3

= 1650 m/s

Velocity of ultrasonics in sea water = 1650 m/s.

 

ASSIGNMENT PROBLEMS

 

1. Calculate the fundamental frequency of 3 mm thick x‒cut quartz crystal. Given Young's modulus for quartz is 7.9 ×10¹⁰ Nm^‒2 and ρ for quartz is 2650 kg m³.  (Ans: 9.09×10⁵ Hz)

 

2. An ultrasonic beam is used to determine the thickness of a steel plate. The velocity of ultrasonics in steel is 5000 m/s. It is noticed that the difference between the two adjacent harmonic frequencies is 56 KHz. Determine the thickness of the given steel plate. (Ans: 0.0446 m)

 

3. Calculate the frequency of the fundamental note and the first overtone emitted by a piezoelectric crystal using the following data. Vibrating length is 4 mm. Young's modulus of quartz = 7.9 × 10¹⁰Nm^‒2 and density of crystal = 2650 kgm^‒3.

(Ans: Fundamental note 6.824 ×10⁵Hz, First overtone = 1.364 ×10⁶Hz)

 

4. Longitudinal standing waves are set up in a quartz plate with antinodes at opposite faces. The fundamental frequency of vibration is given by the relation, f = 2.5 × 10³ / t, where t is the thickness of the plate in metre.

Compute (i) Young's modulus of the quartz plate

(ii) The thickness of the plate required for a frequency of 1400 KHz. The density of quartz is 2,660 kgm^‒3     (E=6.65 ×10¹⁰ Nm^‒2; t=1.785×10^‒3m)