Question

A glider is undergoing simple harmonic motion on a frictionless air track. When it is at position x = 0.30 m, its kinetic energy is half of its potential energy, K=12U. What is the amplitude of its simple harmonic motion?

Answer 1

The amplitude of its simple harmonic motion is about 0.37 m

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Further explanation

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.

$\large{\boxed{T = 2 \pi\sqrt{\frac{m}{k}}}}$

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.

$\large{\boxed{T = 2 \pi\sqrt{\frac{L}{g}}}}$

T = Periode of Pendulum ( second )

L = Length of Pendulum ( kg )

g = Gravitational Acceleration ( m/s² )

Let us now tackle the problem !

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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:

$U_{max} = U + K$

$\frac{1}{2}k A^2 = U + \frac{1}{2}U$

$\frac{1}{2}k A^2 = \frac{3}{2}U$

$\frac{1}{2}k A^2 = \frac{3}{2}( \frac{1}{2}kx^2 )$

$A^2 = \frac{3}{2} x^2$

$A^2 = \frac{3}{2} \times 0.30^2$

$A^2 = 0.135$

$A = \sqrt{0.135}$

$A \approx 0.37 \texttt{ m}$

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Learn more

• 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

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Answer details

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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