How Do Sound Waves Affect Emergency Responses and Wave Motion Calculations?

[Shay10825]

Hi everyone! I need help on these two problems:

1.) A flower pot is knocked off a balcony 20m above the sidewalk and falls toward an unsuspecting 1.75-m-tall man who is standing below. How close to the sidewalk can the flower pot fall before it is too late for a shouted warning from the balcony to reach the man in time? Assume that the man below requires .3 s to respond to the warning.

Would I use the formula:
v=(d)/(delta t)? If so how?
The answer is 7.82m

2. Consider the sinusoidal wave, with the wave function:
y=(15 cm) cos(.157x-50.3t)

At a certain instant, let point A be at the origin and point B be the first point along the x-axis where the wave is 60 degrees out of phase with point A. What is the coordinate of point B?

I did:
y=(15 cm) cos(.157x-50.3t)=60
15cos(.157x)=60
cos(.157x)=4 and this is impossible
I have no clue what to do.

~Thanks

 

[HallsofIvy]

The point of the problem is for the sound to arrive at least .3 s before the pot!

No, you would not use v= d/(delta t). You know the speed of sound, I presume so you can calculate the time required for the sound to travel 20 m. I'm going to call that T. You should replace T by the actual number.
The flower pot starts 20-1.75= 18.25 m above the man's head and has acceleration -9.8 m/s². The height of the flower pot, above the man's head, at time t is 18.25- 4.9t².
If t is the time a which you shout, your shout will arrive at the person's ear in time
t+ T and the person will have moved by time t+ T+ .2. In order that the flower pot have not yet hit the person, we must have
18.25- 4.9(t+ T+ .3)^{2> 0. The "critical time" will be when that is equal to 0.}

 

[HallsofIvy]

The point of the problem is for the sound to arrive at least .3 s before the pot!

No, you would not use v= d/(delta t). You know the speed of sound, I presume so you can calculate the time required for the sound to travel 20 m. I'm going to call that T. You should replace T by the actual number.
The flower pot starts 20-1.75= 18.25 m above the man's head and has acceleration -9.8 m/s². The height of the flower pot, above the man's head, at time t is 18.25- 4.9t².
If t is the time a which you shout, your shout will arrive at the person's ear in time
t+ T and the person will have moved by time t+ T+ .2. In order that the flower pot have not yet hit the person, we must have
18.25- 4.9(t+ T+ .3)<sup>2> 0. The "critical time" will be when that is equal to 0.

As far as 2 is concerned, it makes no sense to set y= 60. y is a height of the wave in cm. "60" is in degrees. Do you understand what "60 degrees out of phase" means? What is it in y= (15 cm)cos(.157x- 50.3t) that IS measured in degrees?</sup>

 

[Galileo]

Hi everyone! I need help on these two problems:

1.) A flower pot is knocked off a balcony 20m above the sidewalk and falls toward an unsuspecting 1.75-m-tall man who is standing below. How close to the sidewalk can the flower pot fall before it is too late for a shouted warning from the balcony to reach the man in time? Assume that the man below requires .3 s to respond to the warning.

Would I use the formula:
v=(d)/(delta t)? If so how?
The answer is 7.82m

if the man needs 0.3 seconds to respond, then the time it takes for the man to react from the moment you warned him is:
Time it takes for the sound to get to him + 0.3 s.
You can calculate the time the sound takes him if you know the speed of sound (something like 330 m/s) and the distance it has to travel (about 20-1.75 m to his ears).
Now calculate how high the flower pot has to be above the man so that it lands on his head presicely when he is about to react. That is the threshold height.

y=(15 cm) cos(.157x-50.3t)=60

This is not correct. A phase difference means the argument of the sine in point B difference from the argument in point A by 60 degrees.