# Pulses in wire - waves

1. Apr 11, 2006

### nick85

Hi, I am having difficulty understanding this question:

A wire 5.0 m long and having a mass of 130 g is stretched under a tension of 220 N. If two pulses, separated in time by 30.0 ms, are generated, one at each end of the wire, where will the pulses first meet?

m (distance from the left end of the wire)

From my deduction the each wave starts from the opposite side.
I figured out the v=91.9866 b/c the square root of 220/(.13/5)=91.9866 m/s

I then set up equations
t=time x=distance
91.9866t=x
91.9866(t-3*10^-3)=5-x

I got x to equal 2.6379799 m
However, I am not certain if this is the correct answer and I am down to my last submission. I was wondering if anyone could agree with what I got.

2. Apr 12, 2006

### Hootenanny

Staff Emeritus
I get 1.12m for my answer. Here's my calcs;

$$v = \sqrt{\frac{T}{\frac{m}{L}}} = \sqrt{\frac{220}{\frac{0.13}{5}}} = 91.9866 m/s$$

$$x = vt \Rightarrow t = \frac{x}{v}$$
$$5 - x = v(t + 0.03)$$

Subbing $t = \frac{x}{v}$ into $5 - x = v(t + 0.03)$ gives;

$$5 - x = v\left( \frac{x}{v} + 0.03 \right) = x + 0.03v$$

$$2x = 5 - 0.03v \Rightarrow x = \frac{5 - 0.03\times 91.9866}{2}$$

$$\fbox{ x = 1.120201m }$$

I'm not sure that I'm right though

-Hoot

3. Apr 12, 2006

### nrqed

I disagree with your answer and agree with Hootenanny. Your mistake is the sign of the 3*10^-3 in the second line. It should be a plus sign.

Basically (after substitution) you wrote x- 91.99* 3*10^-3 = 5 -x or
2x= 5 + 91.99*3*10^-3 or

2x= 5 + 2.76

But it should be 2x= 5 - 2.76.

(it's easy to see. In the first 30 ms, the first pulse travels 2.76 m. There is still a distance of 5-2.76 to travel for both pulses when the second pulse will be emitted. They will obviously meet halfway through this remaining distance, therefore the answer is x= 1/2(5-2.76).

Patrick