# Homework Help: Loudspeaker delay

1. Oct 3, 2014

### toothpaste666

1. The problem statement, all variables and given/known data
For large concerts, loudspeakers are sometimes used to amplify a singer's sound. The human brain interprets sounds that arrive within 50 ms of the original sound as if they came from the same source. Thus if the sound from a loudspeaker reaches a listener first, it would sound as if the loudspeaker is the source of the sound. Conversely, if the singer is heard first and the loudspeaker adds to the sound within 50 ms, the sound would seem to come from the singer, who would now seem to be singing louder. The second situation is desired. Because the signal to the loudspeaker travels at the speed of light 3×108m/s , which is much faster than the speed of sound, a delay is added to the signal sent to the loudspeaker.

How much delay must be added if the loudspeaker is 5.0m behind the singer and we want its sound to arrive 30 ms after the singer's?

2. Relevant equations
D = v/t

3. The attempt at a solution
I am having a hard time figuring out what to do for this problem. I haven't made much progress. If i choose an arbitrary point, the sound from the singer travels a distance d to that point in time t. The sound from the speaker travels distance d + 5m in time t + 30 ms. This doesn't seem like useful information though. Can someone please help me get started?

2. Oct 4, 2014

### toothpaste666

I tried this following solution but it was useless:
the sound from the singer travels distance d where d = vt = 343t (343 is speed of sound in air)
the sound from the speaker travels d + 5 where d + 5 = vt = 343(t + 30) = 343(t-x+30) where x is the delay
when i solve the two equations for x i just end up with 30. I am confusing myself. Can someone give me a hint on how to set this problem up?

3. Oct 4, 2014

### haruspex

I don't understand how you deduced x = 30 from those equations. Maybe your use of t to mean different things in different contexts has confused you.

4. Oct 4, 2014

### toothpaste666

this was my work
d = 343t
d+5 = 343(t-x+30) so d = 343(t-x+30)-5 = 343t - 343x + 343(30) - 5
setting the equations equal to each other
343t = 343t - 343x + 343(30) - 5
0 = -343x + 343(30) -5
343x = 343(30) - 5
343x = 10285
x = 30

so the mistake i made is using the same t for both situations?

5. Oct 5, 2014

### haruspex

Your error is that you have the speed of sound in m/s, but the acceptable delay is in ms, not 30 seconds.

6. Oct 5, 2014

### toothpaste666

ahh so it would be .343x = .343(30)-5
.343x = 5.29
x = .0648 ms

7. Oct 5, 2014

### haruspex

You seem to have made an arithmetic error in the final step. Otherwise correct.

8. Oct 5, 2014

### toothpaste666

oh wow cant believe i did that. thank you!