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
[tex]f = \frac{1}{2\pi }\sqrt{\frac{k}{m}}[/tex]
[tex]f = \frac{1}{2\pi }\sqrt{\frac{2177.78}{8}}[/tex]
[tex]f [/tex] = 2.63 Hz
The frequency of the spring is about 2.6 Hz
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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:
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:
[tex]F = kx[/tex]
[tex]mg = kx[/tex]
[tex]k = mg \div x[/tex]
[tex]k = \frac{mg}{x}[/tex]
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Next, we could calculate the frequency of the spring as follows:
[tex]f = \frac{1}{2 \pi}\sqrt{\frac{k}{m}}[/tex]
[tex]f = \frac{1}{2 \pi}\sqrt{\frac{mg / x}{m}}[/tex]
[tex]f = \frac{1}{2 \pi}\sqrt{\frac{g}{x}}[/tex]
[tex]f = \frac{1}{2 \pi}\sqrt{\frac{9.8}{3.6 \times 10^{-2}}}[/tex]
[tex]f = \frac{35\sqrt{2}}{6\pi} \texttt{ Hz}[/tex]
[tex]f \approx 2.6 \texttt{ Hz}[/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
