Question

An object of mass 8.0 kg is attached to an ideal massless spring and allowed to hang in the Earth's gravitational field. The spring stretches 3.6 cm before it reaches its equilibrium position. If this system is allowed to oscillate, what will be its frequency?

Answer 1

The frequency of the spring is about 2.6 Hz

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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.

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

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.

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

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:

mass of the object = m = 8.0 kg

extension of the spring = x = 3.6 cm = 3.6 × 10⁻² m

Unknown:

frequency of the spring = f = ?

Solution:

Firstly, we will calculate the spring constant as follows:

$F = kx$

$mg = kx$

$k = mg \div x$

$k = \frac{mg}{x}$

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Next, we could calculate the frequency of the spring as follows:

$f = \frac{1}{2 \pi}\sqrt{\frac{k}{m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{mg / x}{m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{g}{x}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{9.8}{3.6 \times 10^{-2}}}$

$f = \frac{35\sqrt{2}}{6\pi} \texttt{ Hz}$

$f \approx 2.6 \texttt{ Hz}$

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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

Answer 2

Answer:

2.63 Hz

Explanation:

m = mass of the object = 8.0 kg

x = stretch in the spring = 3.6 cm = 0.036 m

k = spring constant of the spring

using equilibrium of force

Spring force = weight of object

k x = m g

k (0.036) = (8) (9.8)

k = 2177.78 N/m

frequency of oscillation is given as

$f = \frac{1}{2\pi }\sqrt{\frac{k}{m}}$

$f = \frac{1}{2\pi }\sqrt{\frac{2177.78}{8}}$

$f$ =  2.63 Hz