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Zepler [3.9K]
3 years ago
5

SIMPLE HARMONIC MOTION 1.0 Objective To study simple harmonic motion by observing the motion of a simple pendulum. 2.0 Simple ha

rmonic motion Oscillatory motion is extremely common in nature. Examples include waves (water, sound, earthquake, etc.) and vibrations produced by musical instruments. If the oscillation is characterized by a constant frequency and amplitude (if the motion reproduces itself in a fixed time period T), then the motion is said to be "harmonic." If the oscillation can be described as a sinusoidal function of time and position, the motion is said to be "simple harmonic." Simple harmonic motion (SHM) occurs when for every applied force or torque, there is a restoring force or torque which is proportional to the displacement of the system from its equilibrium position. 2.0.1 Name two real-world examples of harmonic motion. (Hint: Southern California is known for what sort of natural disasters?)
Physics
1 answer:
bagirrra123 [75]3 years ago
8 0

Answer:

the waves in the sea,  leaves of the trees, cables in the bridges, pendulum clock

Explanation:

In nature there are many examples of simple harmonic motion, for example.

* The movement of the waves in the sea is an oscillation movement up and down

* The movement of the leaves of the trees when a wind blows and then stops, but the leaf and branches are oscillating

* The movement of the cables in the bridges, especially in the suspension bridges

* The movement of a pendulum clock

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A 320 g rubber ball is dropped from 2.0 m above the ground, bounces, and returns to a maximum height 1.2 m above the ground. If
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Answer:

0.71 J

Explanation:

320 g = 0.32 kg

According to law of energy conservation, the energy loss to external environment (air, ground) can be credited to the change in mechanical energy of the ball.

As the ball was dropped at H = 2 m above the ground then later reaches its maximum height at h = 1.2m, tts instant speed at those 2 points must be 0. So the kinetic energy at those points are 0 as well. The change in mechanical energy is the change in potential energy.

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What happens to incoming light rays that are parallel to the principal axis of a convex lens?
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PLZ HELP ME!!!!!!!!
mote1985 [20]

Answer:

1.     Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.77955 m

now frequency is given by  

f = \frac{343m/s}{ 0.77955m}

f = 440 Hz  

2.     Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.52028 m

now frequency is given by  

f = \frac{343m/s}{ 0.52028m}

f = 659.3 Hz

3.     Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.65552 m

now frequency is given by  

f = \frac{343m/s}{ 0.65552m}

f = 523.25 Hz  

4.     Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 587.33 Hz

now wavelength is given by  

\lambda = \frac{343m/s}{587.33}

\lambda = 0.584 m    

5.     Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 493.88 Hz

now wavelength is given by  

\lambda = \frac{343m/s}{493.88}

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6.     Wavelength = speed / Frequency

here we know

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now wavelength is given by  

\lambda = \frac{343m/s}{698.46}

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7.     Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.5840 m

now frequency is given by  

f = \frac{343m/s}{0.5840m}

f = 587.3 Hz  

8.     Frequency = speed / wavelength

here we know

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wavelength = 0.4375 m

now frequency is given by  

f = \frac{343m/s}{0.4375m}

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here we know

speed = 343 m/s

Frequency = 783.99 Hz

now wavelength is given by  

\lambda = \frac{343m/s}{783.99}

\lambda = 0.4375 m    

10.  Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 659.26 Hz

now wavelength is given by  

\lambda = \frac{343m/s}{659.26}

\lambda = 0.520 m    

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