How Can I Determine Where the Pulses Will Meet on a Stretched Wire?

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To determine where the pulses will meet on a stretched wire, first calculate the wave speed using the formula c = √(T/ρ), where T is the tension and ρ is the linear density (mass divided by length). For a wire 10.3 m long with a mass of 97.8 g and tension of 248 N, the wave speed is calculated to be 161.6 m/s. Given that the pulses are generated 29.6 ms apart, the distance each pulse travels can be determined using the wave speed and time. The pulses will meet approximately 7.54 m from the starting end of the first pulse. Understanding these calculations allows for accurate predictions of pulse interactions on a stretched wire.
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I'm having a hard time with this problem, and was wondering if somebody can help
A wire 10.3 m long have a mass of 97.8 g is stretched under tension of 248N. If two pulses, separated in time by 29.6 ms, are generated on at each end of the wire, where will the pulses meet?

Plz e-mail if anybody has a way to do this problem
dj_werd@hotmail.com
 
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Unless this is a course in partial differential equations, you were probably given a formula for wave speed on a stretched wire: if I remember correctly it is
c= \sqrt{(\frac{T}{\rho})}
where T is the tension and ρ is the linear density of the wire (mass divided by length). Once you know the wave speed, it should be easy to determine where the pulses meet.
 
I get a wave speed of 161.6 m/s and that the two pulses meet 7.54 m from the end at which the first pulse started.
 
Okay so I understand how to get the wave speed, but I am not really sure how you found the distance on where the pulses will meet.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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