Calculating Doppler Shift with Ship Sonar and Moving Targets

AI Thread Summary
The discussion focuses on calculating the Doppler shift for a ship's sonar system operating at 23.0 kHz with sound traveling in water at 1482 m/s. A whale moving towards the ship at 5.00 m/s alters the frequency of the reflected waves. The participant initially calculated the frequency of the waves reflected from the whale as 23,077 Hz, but encountered an error when determining the frequency difference, arriving at 12 Hz. Key points of confusion included the correct application of the Doppler effect equation and ensuring the wavelength calculation was appropriate for water. Clarifications were provided regarding the reflection frequency and the medium used for wavelength calculations.
jackleyt
Messages
20
Reaction score
0

Homework Statement



The sound source of a ship's sonar system operates at a frequency of 23.0 kHz. The speed of sound in water is 1482 m/s.

What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 5.00 m/s? The ship is at rest in the water.

Homework Equations


fL= fS(v+vL)/(v+vS)


The Attempt at a Solution



I got the first part of the question, which was to find the wavelength. The wavelength is 6.44*10^(-2) m. For the second part, I calculated the frequency of the directly radiated waves and got 23,012.4 Hz. Then, I used the Doppler effect equation and got the frequency that the whale heard, which was 23077 Hz. I used that frequency to determine the reflection, and I got 23,000 Hz. The difference was 12 Hz, which was wrong. Can someone help me with what I'm doing wrong? Thanks!
 
Physics news on Phys.org
A couple of things could have gone wrong here, but from what I see they aren't math related. Either:

1) The frequency of the waves reflected from the whale (the waves that the whale is producing) is 23077 Hz. 23 kHz is the frequency of the waves reflected from the whale and heard by the submarine. Small difference but a difference nonetheless.

2) The wavelength you calculated in a) was in a different medium than water.
 
Thanks! It was the latter of the two.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
2
Views
2K
Replies
7
Views
3K
Replies
1
Views
2K
Replies
1
Views
2K
Replies
4
Views
2K
Replies
5
Views
3K
Back
Top