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liubo4ka [24]
3 years ago
9

A 1.20 m wire has a mass of 6.80 g and is under a tension of 120 N. The wire is held rigidly at both ends and set into oscillati

on. (a) What is the speed of waves on the wire? What is the wavelength of the waves that produce (b) one-loop and (c) two-loop standing waves? What is the frequency of the waves that produce (d) one-loop and (e) two-loop standing waves?
Physics
1 answer:
swat323 years ago
4 0

Answer:

145.52137 m/s

1.4 m

0.7 m

60.6339 Hz

121.2678 Hz

Explanation:

T = Tension = 120 N

\mu = Linear density  = \frac{m}{L}

m = Mass of wire = 6.8 g

L = Length of wire = 1.2 m

n = Number of loops

Velocity is given by

v=\sqrt{\frac{T}{\mu}}\\\Rightarrow v=\sqrt{\frac{T}{\frac{m}{L}}}\\\Rightarrow v=\sqrt{\frac{120}{\frac{6.8\times 10^{-3}}{1.2}}}\\\Rightarrow v=145.52137\ m/s

The speed of waves on the wire is 145.52137 m/s

Wavelength is given by

\lambda=\frac{2L}{n}\\\Rightarrow \lambda=\frac{2\times 1.2}{1}\\\Rightarrow \lambda=1.4\ m

The wavelength of the waves that produces one-loop standing waves is 1.4 m

\lambda=\frac{2L}{n}\\\Rightarrow \lambda=\frac{2\times 1.2}{2}\\\Rightarrow \lambda=0.7\ m

The wavelength of the waves that produces two-loop standing waves is 0.7 m

Frequency is given by

f=\frac{nv}{2L}\\\Rightarrow f=\frac{1\times 145.52137}{2\times 1.2}\\\Rightarrow f=60.6339\ Hz

The frequency of the waves that produces one-loop standing waves is 60.6339 Hz

f=\frac{nv}{2L}\\\Rightarrow f=\frac{2\times 145.52137}{2\times 1.2}\\\Rightarrow f=121.2678\ Hz

The frequency of the waves that produces two-loop standing waves is 121.2678 Hz

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and it won't matter what route the tool follows to get anywhere. 
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Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for  Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

        Work  =  integral of (F·dy) evaluated from  0  to  2.40

                  =  integral of (-2.85 y² dy) evaluated from  0  to  2.40

                 =  (-2.85) · integral of  (y² dy)  evaluated from  0  to  2.40 .


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Evaluated from  0  to  2.40 , it's  (1/3 · 2.40³) - (1/3 · 0³)

                                            =  1/3 · 13.824  =  4.608 .

And the work  =  (-2.85) · the integral

                     =  (-2.85) · (4.608)

                     =      - 13.133  .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
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-- The work done by the force is negative, because the force points
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itself did 13.133 of negative work to 'allow' the tool to move up. 

-- It doesn't matter whether the tool goes there along the line  x=y , or
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