The Doppler Effect and Simple Harmonic Motion question

• hey3796
In summary: :) in summary, the ultrasonic motion detector can be used to detect the fetus's heartbeat by measuring the amplitude and frequency of the wall's motion.
hey3796

Homework Statement

Ok here's the problem.. a little lengthy so bear with me :) (3 parts need help w/ last two)

Expectant parents are thrilled to hear their unborn baby's heartbeat, revealed by an ultrasonic motion detector. Suppose the fetus's ventricular wall moves in simple harmonic motion with an amplitude of 1.75 mm and a frequency of 125 per minute.

(a) Find the maximum linear speed of the heart wall. Suppose the motion detector in contact with the mother's abdomen produces sound at 2,000,000.0 Hz which travels through tissue at 1.50 km/s.

(b) Find the difference between the original frequency and the maximum frequency at which sound arrives at the wall of the baby's heart.

(c) Find the difference between the original frequency and the maximum frequency at which reflected sound is received by the motion detector. By electronically "listening" for echoes at a frequency different from the broadcast frequency, the motion detector can produce beeps of audible sound in synchronization with the fetal heartbeat.

Homework Equations

ω=2pi(f) where w is angular freq. and f is freq
vmax=ωA
f'= (v+vo/v) and f"= (v/v-vs) where vs is the speed of sound

The Attempt at a Solution

Ok so for part a) I determined Vmax by using Vmax=wA. No problem there. cool.

Parts b) and c): this is where it gets really frustrating!

For part b) I thought ok were doing the doppler effect. The question asks for the DIFFERENCE between FREQUENCIES. Ok, easy enough. NOT

The answer is going to be in HZ. The sound travels through the tissue @ 1.50 km/s @ a frequency of 2000000.0 HZ and a max speed of .022925 from part a).

I used the equation f'= f(v+vo/v)= 2000000.0Hz((1500m/s + .022925m/s)/(1500m/s).
Right? WRONG!

Plz help me... I am really stumped. What am I missing here?

so you say that

MaxVelocity = 2*PI*Frequency*Amplitude

so that would be

2*3.14*(125/sec)*0.00175metres

I don't get the same figure as you do

i didnt have any issues with that part.

I was having problems with the last two parts but i figured it out. I simply forgot to subtract the answer i got from the original frequency (2000000 HZ)

SOLVED!

What is the Doppler Effect?

The Doppler Effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. It is commonly observed with sound waves, such as the change in pitch of a siren as an ambulance drives by.

How does the Doppler Effect relate to Simple Harmonic Motion?

The Doppler Effect is closely related to Simple Harmonic Motion because both involve the periodic motion of a wave. Simple Harmonic Motion is a specific type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium. The Doppler Effect involves a change in frequency or wavelength due to the motion of the wave source or observer, which can also be described as a periodic motion.

What is the equation for the Doppler Effect?

The equation for the Doppler Effect is f' = f((v ± vo) / (v ± vs)), where f is the original frequency, f' is the observed frequency, v is the speed of the wave, vo is the velocity of the observer, and vs is the velocity of the wave source. The plus or minus sign depends on whether the observer is moving towards or away from the source.

How is the Doppler Effect used in real life?

The Doppler Effect is used in many real-life applications, such as weather forecasting, astronomy, and medical imaging. In weather forecasting, the Doppler Effect is used to measure the speed and direction of wind using a Doppler radar. In astronomy, it is used to study the motion of stars and galaxies and determine their velocities. In medical imaging, the Doppler Effect is used to measure blood flow and detect abnormalities in the body.

What are some limitations of the Doppler Effect?

The Doppler Effect has some limitations, one of which is that it only applies to waves with a constant frequency. It also assumes that the wave source and observer are moving in a straight line and that the speed of the wave is constant. Additionally, the Doppler Effect is only accurate for small velocities compared to the speed of the wave. For larger velocities, the equations become more complex and may not accurately represent the observed frequency shift.

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