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SCORPION-xisa [38]
3 years ago
9

What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? Vie

w Available Hint(s) What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude.
Physics
2 answers:
ozzi3 years ago
6 0

Answer:

It has no effect on the amplitude.

Explanation:

When the sandbag is dropped, then the cart is at its maximum speed. Dropping the sand bag does not affect the speed instantly, this is because the energy remains within the system after the bag as been dropped. The cart will always return to its equilibrium point with the same amount of kinetic energy, as a result the same maximum speed is maintained.

amm18123 years ago
5 0

Answer:

It decreases the amplitude.

Explanation:

If it is located at the end, its kinetic energy will be equal to zero, so the system as a whole would only have potential energy, which is defined as:

Ep = (1/2) * k * A²

Where A is the amplitude of the oscillation.

According to the energy principle, said energy must remain conserved, therefore, the total energy is equal to:

Etot = (1/2) * k * A²

Since the system is in equilibrium, according to the expression of conservation of energy, the energy in the position located at the end would be equal to the energy in equilibrium. As the sandbag falls, its kinetic energy decreases, which is why the total energy of the system also decreases.

According to the total energy expression, it is directly proportional to the square of the oscillation amplitude. If the total energy is decreased, the amplitude of the oscillation would also decrease.

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A mass weighing 14 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping force numeri
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The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

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Explanation:

Using the newton second law

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m mass attached

m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}

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\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0

Converting units of weights in units of mass (equation of motion)

m = \frac{W}{g} = \frac{14}{32} = 0.43 slug

From hook's law we can calculate the spring constant k

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If we put m and k into the DE, we get

\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0

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2λ = \frac{b}{m} = \frac{b}{0.43}

λ = b/0.215

w^{2} = \frac{k}{m} = 16.28

λ^2 - w^2 = \frac{b^{2} }{0.046} - 16.28

This way,

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

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