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
[tex]v_{max}=2\pi f A[/tex]
[tex]A=\dfrac{v}{2\pi f}[/tex]
Where,
v = speed
f = frequency
A = amplitude
Put the value into the formula
[tex]A=\dfrac{2.5}{2\pi\times250}[/tex]
[tex]A=0.00159 = 1.59\times10^{-3}\ m[/tex]
[tex]A=1.6\ mm[/tex]
Hence, The amplitude of the wing tip's motion is 1.6 mm.
The amplitude of the wing tip's motion is about 1.6 mm
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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.
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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.
[tex]T = 2 \pi\sqrt{\frac{m}{k}}[/tex]
T = Periode of Spring ( second )
m = Load Mass ( kg )
k = Spring Constant ( N / m )
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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.
[tex]T = 2 \pi\sqrt{\frac{L}{g}}[/tex]
T = Periode of Pendulum ( second )
L = Length of Pendulum ( kg )
g = Gravitational Acceleration ( m/s² )
Let us now tackle the problem !
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Given:
frequency of the wings = f = 250 Hz
maximum speed of the wings = v_max = 2.5 m/s
Asked:
amplitude of the wing tip's motion = A = ?
Solution:
[tex]v = \omega \sqrt{A^2 - x^2}[/tex]
[tex]v_{max} = \omega \sqrt{A^2 - 0^2}[/tex]
[tex]v_{max} = \omega A[/tex]
[tex]v_{max} = 2 \pi f A[/tex]
[tex]2.5 = 2 \pi \times 250 \times A[/tex]
[tex]2.5 = 500 \pi \times A[/tex]
[tex]A = 2.5 \div (500 \pi)[/tex]
[tex]A \approx 1.6 \times 10^{-3} \texttt{ m}[/tex]
[tex]A \approx 1.6 \texttt{ mm}[/tex]
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Grade: High School
Subject: Physics
Chapter: Simple Harmonic Motion
