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Sunny_sXe [5.5K]

3 years ago

11

A spring is attached to the ceiling and pulled 8 cm down from equilibrium and released. The amplitude decreases by 17% each seco

nd. The spring oscillates 19 times each second. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

1 answer:

Vadim26 [7]3 years ago

3 0

Answer:

D(t) = 8(0.83)^(t) cos 38πt

Explanation:

We are told that the spring oscillates 19 times each second.

Thus, period = 1/19

We are also told that it's pulled 8cm downwards and the amplitude decreases by 17% each second.

Thus;

Amplitude;A = 8 × (1 - (17/100))^(t)

A = 8(0.83)^(t)

If we consider the function;

y = A cos (bx - c) + d

Now, 2π/b = period

So, 1/19 = 2π/b

b = 38π

So, D(t) = 8(0.83)^(t) cos (38πt - c) + d

Since we started from minimum,

Vertical shift, d = 0 and horizontal shift c = 0

So,we now have;

D(t) = 8(0.83)^(t) cos 38πt

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a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

a)

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

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b)

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

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Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

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The vertical speed instead is given by the suvat equation:

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d)

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$v_x=d\sqrt{\frac{g}{2h}}$ is the horizontal component (in the forward direction)

$v_y=\sqrt{2gh}$ is the vertical component (in the downward direction)

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So

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