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With each beat of your heart the aortic valve opens and closes. The valve opens and closes very rapidly, with a peak velocity as high as 4 m/s. If we image it with 7 MHz sound and the speed of sound is approximately 1500 m/s in human tissue, what is the frequency shift between the opening and closing of the valve?

Respuesta :

Answer:

|Δf| = 37.3 kHz

Explanation:

given,

peak velocity = 4 m/s

speed of the sound = 1500 m/s

frequency = 7 MHz

[tex]v = C\dfrac{\pm \dlta f}{2 f_0}[/tex]

[tex]\delta f = \pm 2 f_0 (\dfrac{V}{C})[/tex]

[tex]\delta f = \pm 2\times 7 (\dfrac{4}{1500})[/tex]

           [tex]=\pm 0.0373 MHz[/tex]

           = 37.3 kHz

|Δf| = 37.3 kHz

hence, frequency shift between the opening and closing valve is 37.3 kHz

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