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attashe74 [19]
3 years ago
6

A standing wave with a fundamental frequency 1.25x10 2 Hz is set up on a string of length 0.750 meters. What is the frequency of

the third harmonic?
Physics
1 answer:
Nezavi [6.7K]3 years ago
8 0

Answer:

<em>500Hertz</em>

Explanation:

Given the fundamental frequency of the string given as  1.25x10²Hz

Harmonics are integral multiples of the fundamental frequency.

In strings, the third harmonic F3 is expressed in terms of the fundamental frequency expressed as;

F3 = 4F0

F0 is the fundamental frequency;

Given

F0 = 1.25x10²Hz

F3 = 4(1.25x10²)

F3 = 4(125)

F3 = 500Hz

<em>Hence the  frequency of the third harmonic is 500Hertz</em>

<em></em>

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The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

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a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma   (3)

-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma   (4)

If we know that \theta = 60^{\circ}, \gamma = 19.889^{\circ}, r_{BD} = 10\,in, \omega_{AB} = 16\,\frac{rad}{s}, r_{AB} = 3\,in and \alpha_{AB} = 0\,\frac{rad}{s^{2}}, then the solution of the systems of linear equations are, respectively:

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v_{D} = -32.887\,\frac{in}{s}, \omega_{BD} = -2.552\,\frac{rad}{s}

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a_{D} = -693.867\,\frac{in}{s^{2}}, \alpha_{BD} = 73.082\,\frac{rad}{s^{2}}

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<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

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