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Punkyc7
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A 1800 N irregular beam is hanging by its ends from a ceiling by 2 vertical wires A and B. Each are 1.24 m long and they both weigh 2.9 N. The center of gravity of this beam is 1/3 of the way along the beam from the end where wire A is attached.
If you pluck both strings at the same time at the bottom, what is the time delay between the arrival of the 2 pulses at the ceiling
v= sqrt(T/mu)
so for the velocity in strng A I did va=sqrt((1/3T+5.8)/(.23864)=50.22
did the same thing for b and got 71.026
so for the time we do Deltax=vt so t=1.24/v
ta=.0246
tb= .0174
to the time difference is .00714
my question is if I did everything right?
since the tension is higher in beam A shouldn't that one reach the ceiling for B, for some reason it turned out the other way around when I solved for v
If you pluck both strings at the same time at the bottom, what is the time delay between the arrival of the 2 pulses at the ceiling
v= sqrt(T/mu)
so for the velocity in strng A I did va=sqrt((1/3T+5.8)/(.23864)=50.22
did the same thing for b and got 71.026
so for the time we do Deltax=vt so t=1.24/v
ta=.0246
tb= .0174
to the time difference is .00714
my question is if I did everything right?
since the tension is higher in beam A shouldn't that one reach the ceiling for B, for some reason it turned out the other way around when I solved for v
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