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RideAnS [48]
3 years ago
15

A flagpole consists of a flexible, 7.107.10 m tall fiberglass pole planted in concrete. The bottom end of the flagpole is fixed

in position, but the top end of the flagpole is free to move. What is the lowest frequency standing wave that can be formed on the flagpole if the wave propagation speed in the fiberglass is 27302730 m/s?
Physics
1 answer:
vladimir1956 [14]3 years ago
3 0

To develop this problem we require the concepts related to Frequency and their respective way of calculating it.

The formula to calculate the frequency is given by

f=\frac{V}{\lambda}

Where,

\lambda = wavelength

V= velocity

For fundamental mode, wavelength is equal to 4 time the length.

Then,

\lambda = 4L = 4*7.10m=28.4m

Replacing in the first equation,

f=\frac{2730}{28.4}\\

f= 96.12Hz

Therefore the frequency is 96.12Hz

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Answer:

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Explanation:

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Generally the moment of inertia of one side of the square is mathematically represented as

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Now substituting for  m_1

  So

       I _g=  \frac{1}{12}  *  \frac{M}{4} * a^2

Now according to  parallel-axis theorem the moment of inertia of one side of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

      I_a =  I_g + m [\frac{q}{2} ]^2

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=>    I_a =  \frac{1}{12}  *  \frac{M}{4} * a^2 + {\frac{M}{4} }* [\frac{q}{2} ]^2

=>    I_a = \frac{Ma^2}{48} + \frac{Ma^2}{16}

=>    I_a = \frac{Ma^2}{12}

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=>   I_s = 4 * \frac{Ma^2}{12}

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