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irina1246 [14]
3 years ago
8

We can model the motion of a bumblebee's wing as simple harmonic motion. A bee beats its wings 250 times per second, and the win

g tip moves at a maximum speed of 2.5 m/s . Part A What is the amplitude of the wing tip's motion?

Physics
2 answers:
zlopas [31]3 years ago
3 0

Answer:

The amplitude of the wing tip's motion is 1.6 mm.

Explanation:

Given that,

Beat = 250 /s

Speed = 2.5 m/s

We need to calculate the amplitude of the wing tip's motion

Using the equation for the maximum velocity

v_{max}=2\pi f A

A=\dfrac{v}{2\pi f}

Where,

v = speed

f = frequency

A = amplitude

Put the value into the formula

A=\dfrac{2.5}{2\pi\times250}

A=0.00159 = 1.59\times10^{-3}\ m

A=1.6\ mm

Hence, The amplitude of the wing tip's motion is 1.6 mm.

ASHA 777 [7]3 years ago
3 0

The amplitude of the wing tip's motion is about 1.6 mm

\texttt{ }

<h3>Further explanation</h3>

Simple Harmonic Motion is a motion where the magnitude of acceleration is directly proportional to the magnitude of the displacement but in the opposite direction.

\texttt{ }

The pulled and then released spring is one of the examples of Simple Harmonic Motion. We can use the following formula to find the period of this spring.

T = 2 \pi\sqrt{\frac{m}{k}}

T = Periode of Spring ( second )

m = Load Mass ( kg )

k = Spring Constant ( N / m )

\texttt{ }

The pendulum which moves back and forth is also an example of Simple Harmonic Motion. We can use the following formula to find the period of this pendulum.

T = 2 \pi\sqrt{\frac{L}{g}}

T = Periode of Pendulum ( second )

L = Length of Pendulum ( kg )

g = Gravitational Acceleration ( m/s² )

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

frequency of the wings = f = 250 Hz

maximum speed of the wings = v_max = 2.5 m/s

<u>Asked:</u>

amplitude of the wing tip's motion = A = ?

<u>Solution:</u>

v = \omega \sqrt{A^2 - x^2}

v_{max} = \omega \sqrt{A^2 - 0^2}

v_{max} = \omega A

v_{max} = 2 \pi f A

2.5 = 2 \pi \times 250 \times A

2.5 = 500 \pi \times A

A = 2.5 \div (500 \pi)

A \approx 1.6 \times 10^{-3} \texttt{ m}

A \approx 1.6 \texttt{ mm}

\texttt{ }

<h3>Learn more</h3>
  • Model for Simple Harmonic Motion : brainly.com/question/9221526
  • Force of Simple Harmonic Motion : brainly.com/question/3323600
  • Example of Simple Harmonic Motion : brainly.com/question/11892568

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Simple Harmonic Motion

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Answer:

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Explanation:

Given that

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Rotation energy ,RE

RE=\dfrac{1}{2}I\omega^2

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RE=\dfrac{1}{5}\times MR^2\times \omega^2

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Explanation:

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