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sergeinik [125]
4 years ago
11

Edit question A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a t

ube of length 192 cm that is open near the wire and closed at the other end. Please find the tension in the wire assuming that the speed of sound in air is 343 m/s.
Physics
1 answer:
Pachacha [2.7K]4 years ago
7 0

Answer:

176.9N

Explanation:

The following data were given

wire length,L=37cm=0.37m

linear density=18g/m

tube length,=192cm=1.92m,

speed of sound,v=343m/s

Since it is an open-closed tube, the second harmonic frequency is expressed as

f_{3}=3(\frac{v}{4l} )\\f_{3}=3(\frac{343}{4*1.92})\\f_{3}=133.98Hz

The relationship between the tension,  linear density and second harmonic frequency is expressed as

f_{3}=\frac{1}{2l_{w}}\sqrt{\frac{T}{\alpha } } \\T=(f_{3}*2l_{w})^{2}\alpha \\T=(133.984*2*0.37)^{2}*18*10^{-3}\\T=176.9N

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Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

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Answer:

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Explanation:

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    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

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      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

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From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

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         Q =2.094 C

     

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