Diagnostic ultrasound of frequency 3.45 MHz is used to examine tumors in soft tissue.

(a) What is the wavelength in air of such a sound wave?
(b) If the speed of sound in tissue is 1500 m/s, what is the wavelength of this wave in tissue? (Take the speed of sound in air to be 343 m/s.)

4.35×10⁻⁴ m

Explanation:

(a)

From wave,

v = λf Equation 1

Where v = speed of sound in air, λ = wavelength of sound, f = frequency of sound.

Make λ the subject of formula in  equation 1

λ = v/f Equation 2

Given: v = 343, f = 3.45 MHz = 3.45×10⁶ Hz

Substitute into equation 2

λ = 343/(3.45×10⁶)

λ = 99.42×10⁻⁶ m

λ = 9.942×10⁻⁵ m

(b)

using,

v' = λ'f Equation 3

Where v' = speed of sound in tissue, λ' = wavelength of sound in tissue.

make λ' the subject of the equation

λ' = v'/fEquation 4

Given: v' = 1500 m/s, f = 3.45 MHz = 3.45×10⁶ Hz

Substitute into equation 4

λ' = 1500/(3.45×10⁶)

λ' = 434.783×10⁻⁶

λ' ≈ 4.35×10⁻⁴ m.

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.