Doppler Effect with heart motion

[rpcarroll]

Homework Statement

Suppose a heart's ventricular wall moves in simple harmonic motion with amplitude 1.8mm and a frequency of 115 per minute. (a) Given a motion detector in contact with the chest produces sound at 2,000,000Hz which travels through tissue at 1.50 km/s. Find the maximum linear speed of the heart wall. (b) Find maximum frequency at which the sound arrvies at the heart. (c) Find the maximum frequency which at which sound is received by the motion detector.

Homework Equations

f'=fo((Vo+Vmax)/Vo)

The Attempt at a Solution

(a) 115=2,000,000((1.5+Vmax)/1.5)
Vmax=1.5
I am struggling with some of the concepts and where different terms can be applied, any suggestions? Do I need to convert any units?

 

[turin]

... simple harmonic motion with amplitude 1.8mm and ... frequency of 115 per minute.
(a) Given ... sound at 2,000,000Hz ... travels through tissue at 1.50 km/s. Find the maximum linear speed of the heart wall. (b) Find maximum frequency at which the sound arrvies at the heart. (c) Find the maximum frequency ... at which sound is received by the motion detector.

The wording of (a) is quite confusing, but I think that you don't need the info about the motion detector. (b) is ambiguous, but I would imagine you need to take into acount the doppler effect. You will have to think about (c) a bit.

(a) 115=2,000,000((1.5+Vmax)/1.5)
Vmax=1.5

I don't think so.

 

[thomate1]

I would suggest breaking down the problem into smaller, more manageable parts and using the relevant equations and concepts to solve each part.

First, let's focus on part (a). We are given the amplitude (1.8mm) and frequency (115 per minute) of the heart's motion, as well as the speed of sound through tissue (1.50 km/s) and the frequency of the sound produced by the motion detector (2,000,000Hz). Our goal is to find the maximum linear speed of the heart wall.

We can use the equation for simple harmonic motion, x = A*cos(2πft), where x is the displacement, A is the amplitude, f is the frequency, and t is time. Since we are given the amplitude and frequency, we can plug them in and solve for the displacement at any given time.

In this case, we are interested in the maximum displacement, which occurs at t = 0. This means that the maximum linear speed of the heart wall is equal to the amplitude, 1.8mm.

Now, let's move on to part (b). We are asked to find the maximum frequency at which the sound arrives at the heart. This is a bit trickier, as it involves the Doppler effect. We can use the equation f' = f₀((v₀ + v)/v₀), where f' is the observed frequency, f₀ is the emitted frequency, v₀ is the speed of the sound source, and v is the speed of the observer.

In this case, the sound source is the motion detector, which is emitting sound at a frequency of 2,000,000Hz. The speed of the sound source is the speed of sound through tissue, 1.50 km/s. The observer is the heart, which is moving towards the sound source at a maximum linear speed of 1.8mm. We can convert this to meters per second (m/s) by dividing by 60 seconds (since the frequency is given in minutes). This gives us a maximum linear speed of 0.03 m/s.

Plugging these values into the equation, we get f' = 2,000,000((1.50 + 0.03)/1.50) = 2,020,000Hz. This is the maximum frequency at which the sound arrives at the heart.

Finally, for part (c), we are asked to find