The amplitude of its simple harmonic motion is about 0.37 m
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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]\large{\boxed{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]\large{\boxed{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:
displacement = x = 0.30 m
kinetic energy = ½ × potential energy
Asked:
amplitude = A = ?
Solution:
We will use conservation of energy as follows:
[tex]U_{max} = U + K[/tex]
[tex]\frac{1}{2}k A^2 = U + \frac{1}{2}U[/tex]
[tex]\frac{1}{2}k A^2 = \frac{3}{2}U[/tex]
[tex]\frac{1}{2}k A^2 = \frac{3}{2}( \frac{1}{2}kx^2 )[/tex]
[tex]A^2 = \frac{3}{2} x^2[/tex]
[tex]A^2 = \frac{3}{2} \times 0.30^2[/tex]
[tex]A^2 = 0.135[/tex]
[tex]A = \sqrt{0.135}[/tex]
[tex]A \approx 0.37 \texttt{ m}[/tex]
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Grade: High School
Subject: Physics
Chapter: Simple Harmonic Motion
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Keywords: Simple , Harmonic , Motion , Pendulum , Spring , Period , Frequency
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