An ultrasonic ruler, such as the one discussed in Example 4 in Section 16.6, dis

Question

An ultrasonic ruler, such as the one discussed in Example 4 in Section 16.6, displays the distance between the ruler and an object, such as a wall. The ruler sends out a pulse of ultrasonic sound and measures the time it takes for the pulse to reflect from the object and return. The ruler uses this time, along with a preset value for the speed of sound in air, to determine the distance. Suppose that you use this ruler under water, rather than in air. The actual distance from the ultrasonic ruler to an object is 24.0 m. The adiabatic bulk modulus and density of seawater are Bad = 2.37 × 109 Pa and = 1025 kg/m3, respectively. Assume that the ruler uses a preset value of 343 m/s for the speed of sound in air. Determine the distance reading that the ruler displays.

Explanation / Answer

Speed of Sound in fluids.

c = (E / )1/2   

where

E = adiabatic bulk modulus (Pa, psi) , 2.37 × 10^9 Pa

= density (kg/m3, lb/ft3) , 1025 kg/m3

c= 1520.5 m/s

--------

The actual time taken by sound wave to travel 48 m (24+ 24) is T=Distance/Speed = 48/1520.5 = 0.031 sec.

But the ruler uses this time, along with a preset value for the speed of sound in air which is 343 m/s to find the distance.

So , the distance reading that the ruler displays will be distance travelled in 0.031sec / 2 with speed 343m/s

=0.031 * 343/ 2 = 5.414 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.