
#1
Apr1106, 08:19 PM

P: 6

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(t3*10^3)=5x 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
Apr1206, 05:18 AM

Emeritus
Sci Advisor
PF Gold
P: 9,789

I get 1.12m for my answer. Here's my calcs;
[tex]v = \sqrt{\frac{T}{\frac{m}{L}}} = \sqrt{\frac{220}{\frac{0.13}{5}}} = 91.9866 m/s[/tex] [tex]x = vt \Rightarrow t = \frac{x}{v}[/tex] [tex]5  x = v(t + 0.03)[/tex] Subbing [itex]t = \frac{x}{v}[/itex] into [itex]5  x = v(t + 0.03)[/itex] gives; [tex]5  x = v\left( \frac{x}{v} + 0.03 \right) = x + 0.03v[/tex] [tex]2x = 5  0.03v \Rightarrow x = \frac{5  0.03\times 91.9866}{2}[/tex] [tex]\fbox{ x = 1.120201m }[/tex] I'm not sure that I'm right though Hoot 



#3
Apr1206, 05:39 AM

Sci Advisor
HW Helper
P: 2,886

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 52.76 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(52.76). Patrick 


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