Q81PP - University Physics with Modern Physics

Question

For cranial ultrasound, why is it advantageous to use frequencies in the $\mathbf{kHz}$  range rather than the $\mathbf{MHz}$ range?

Step-by-Step Solution

Answer

The correct option is  there will be no standing waves at the lower frequencies.

Step 1: Nodes and Antinodes

The antinodes are formed at points of constructive interference and the nodes are formed at points of destructive interference.

The distance between the node is related to the wavelength as

 $\mathbf{d}\mathbf{=}\frac{\mathrm{n\lambda }}{2}\mathbf{=}\frac{\mathrm{nv}}{2f}$

Step 2: calculation

As shown by this equation, the distance  $d$  is inversely proportional to the frequency $f$ . So, as   $f$decreases, the distance increases and vice versa. Therefore, forming waves larger than $0.77 mm$  is difficult. So, using $1kHZ$ frequencies is better than using  $1MHz$.

Hence, there will be no standing waves at the lower frequencies.