Question

A physician is making an image with ultrasound of initial intensity 1000 W/m². When the frequency is set to 1.0 MHz, the intensity drops to 500 W/m² at a certain depth in the patient’s body. What will be the intensity at this depth if the physician changes the frequency to 2.0 MHz?

A. 750 W/m²
B. 500 W/m²
C. 250 W/m²
D. 125 W/m²

Answer 1

(C) The intensity will drop to 250 W/m² when the frequency is changed to 2.0 MHz at this particular depth.

Step-by-step explanation:

Sound intensity is directly proportional to its frequency.

F ∝ I

Sound intensity is inversely proportional to square of distance

I ∝1/r²,

Therefore, F ∝ 1/r²

From the question we were given the following;

F₁ initial frequency = 1.0MHz

I₁ initial intensity = 1000 W/m₂

F₂ Final frequency = 2.0 MHz

When the frequency was set to 1 MHz, the intensity dropped to 500 W/m² at this certain depth.

Hence the decreasing factor = 1000/500 = 2

When the frequency is set to 2 MHz, the depth will drop by 2²

The new intensity = 1000/4 = 250 W/m²

Therefore, the intensity will drop to 250 W/m² when the frequency is changed to 2.0 MHz at this particular depth.