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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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Starting with an initial speed of 4.54 m/s at a height of 0.337 m, a 2.03-kg ball swings downward and strikes a 4.97-kg ball tha
Vsevolod [243]

Answer:

a) 5.2170 m/s

b) 2.1911 m/s, to the left

c) 3.0259 m/s, to the right

d) 0.2449 m

Explanation:

Given parameters:

v_1 = 4.54 m/s

h_1 = 0.337 m

m_1 = 2.03 kg

m_2 = 4.97 kg

v_2 = 0 m/s

a) Here, the conservation of energy principle can be used.

KE_i + PE_i = KE_f + PE_f\\\frac{1}{2}m_1v_1^2+m_1gh_1 = \frac{1}{2}m_1v_{1f}^2+m_1gh_{1f} where h_{1f} = 0

So,

\frac{1}{2}v_1^2+gh_1 = \frac{1}{2}v_{1f}^2\\v_{1f} = \sqrt{2(\frac{1}{2}v_1^2+gh_1)} = \sqrt{2(\frac{1}{2}*4.54^2+9.8*0.337)} = 5.2170 m/s

b) Here, the conservation of momentum principle can be used.

v_{1f} = \frac{m_1-m_2}{m_1+m_2}*v_1

Here, our new v_1 will be  v_{1f} found in part (a).

v_{1f} = \frac{2.03-4.97}{2.03+4.97} * 5.217 = -2.1911 m/s

(-) sign means that the 2.03 kg ball after collision will bounce back in the left direction.

c) Here, we can use the conservation of momentum principle again.

v_{2f} = \frac{2m_1}{m_1+m_2}*v_1. Here, v_1 will be  v_{1f} found in part (a) again.

v_{2f} = \frac{2 * 2.03}{2.03+4.97} * 5.217 = 3.0259 m/s

(+) sign means that the 4.97 kg ball after collision will swing to the right direction.

d) Here, we can use the conservation of energy principle again.

\frac{1}{2}mv^2+mgh = \frac{1}{2}mv_f^2+mgh_f where h = 0 and v_f = 0

So,

gh_f = \frac{1}{2}v_1^2

h_f = \frac{v_1^2}{2g} = \frac{(-2.1911)^2}{2*9.8} = 0.2449 m

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Which simple machine is NOT correctly matched with an appropriate task for its use?
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Block m1 of mass 2m and velocity v0 is traveling to the right (+x) and makes an elastic head-on collision with block m2 of mass
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