Question

7. 0.41 points Tipler8 2.P057 My Notes Ask Your Teacher A submarine can use sonar (sound traveling through water) to determine its distance from other objects. The time between the emission of a sound pulse (a "ping") and the detection of its echo can be used to determine such distances. Alternatively, by measuring the time between successive echo receptions of a regularly timed set of pings, the submarine's speed may be determined by comparing the time between pings. Assume you are the sonar operator in a submarine traveling at a constant velocity underwater. Your boat is in the eastern Mediterranean Sea, where the speed of sound is known to be 1522 m/s. If you sent out pings every 5.40 s, and your apparatus receives echoes reflected from an undersea cliff every 5.37 s, how fast is your submarine approaching the cliff? m/s eBook

7,a,b,c,d please

Answer 1

Given is:-

Speed of the sound is = 1522 m/s

time ping sent out at = 5.40 seconds

time in receiving echoes = 5.37 seconds

Now,

This problem takes advantage of the fact that both the emitted and reflected pings are equally spaced with distance S, and traveling at speed vsound = 1522 m/s. We will first determine the spacing of the sub. At time t=0, the submarine emits the first ping from the origin, the ping travels to the right with speed V_sound, so at time t=T=5.4 s, it is at position

$d_1=V_{sound} \times Time = V_{sound}T$

at time t=T, the submarie is now at location

$d_2=V_{sub} \times Time = V_{sub}T$

and since the sub now emits a second ping, at t=T, ping 2 is at

$V_{sub}T$

The pings are traveling spaced by distance S given by

$S =d_1-d_2$

$S = V_{sound}T - V_{sub}T$ eq-1

now we look at the situation after reflection. The pings are traveling in the -x direction while the sub continues to travel in the +x direction. At time t=0, ping 1 reaches the sub, and clearly ping 2 is now at x=+S. At time t=t2=5.37s, the sub meets ping2. Asssuming the position of the sub at t=0 to be the origin, then the position of the sub at t=t₂ is given by

$X_s = V_{sub}t_2$

ping 2, however, started at S, and travels in the -x direction with speed vsound, so its osition at t=t₂ is

$X_p = S - V_{sound}t_2$

by putting the value of S from eq-1 we get

$X_p = -V_{sub}T + V_{sound}T - V_{sound}t_2$

The two meet at t=t2, so xp = xs, and we solve for vsub

$V_{sub} = \frac{V_{sound}T - t_2 V_{sound}}{T+t_2}$

by putting all the values in abvoe equation, we get

$V_{sub} = \frac{(1522)(5.4) - (5.37)(1522)}{5.40+5.37}$

$V_{sub} = 4.24m/s$

Therefore the submarine is approaching the cliff at a speed of 4.24m/s