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Vlad1618 [11]
3 years ago
11

A spring attached to the ceiling is stretched one foot by a four pound weight. The mass is set in motion by pulling down 2 feet

and then released in a medium with a damping force numerically equal to the velocity.
a. Find the Hook's law spring constant k.
b. Find its natural frequency ω.
c. Form a differential equation for the position x(t) of the mass.
d. Determine the solution for the position (in alternate form).
e. Find the times at which the mass passes the equilibrium second time heading up.
Physics
1 answer:
Marina CMI [18]3 years ago
5 0

Given that,

Weight = 4 pound

W=4\ lb

Stretch = 2 feet

Let the force be F.

The elongation of the spring after the mass attached is

x=2-1=1\ feet

(a). We need to calculate the value of spring constant

Using Hooke's law

F=kx

k=\dfrac{F}{x}

Where, F = force

k = spring constant

x = elongation

Put the value into the formula

k=\dfrac{4}{1}

k=4

(b). We need to calculate the mass

Using the formula

F=mg

m=\dfrac{F}{g}

Where, F = force

g = acceleration due to gravity

Put the value into the formula

m=\dfrac{4}{32}

m=\dfrac{1}{8}\ lb

We need to calculate the natural frequency

Using formula of natural frequency

\omega=\sqrt{\dfrac{k}{m}}

Where, k = spring constant

m = mass

Put the value into the formula

\omega=\sqrt{\dfrac{4}{\dfrac{1}{8}}}

\omega=\sqrt{32}

\omega=4\sqrt{2}

(c). We need to write the differential equation

Using differential equation

m\dfrac{d^2x}{dt^2}+kx=0

Put the value in the equation

\dfrac{1}{8}\dfrac{d^2x}{dt^2}+4x=0

\dfrac{d^2x}{dt^2}+32x=0

(d). We need to find the solution for the position

Using auxiliary equation

m^2+32=0

m=\pm i\sqrt{32}

We know that,

The general equation is

x(t)=A\cos(\sqrt{32t})+B\sin(\sqrt{32t})

Using initial conditions

(I). x(0)=2

Then, x(0)=A\cos(\sqrt{32\times0})+B\sin(\sqrt{32\times0})

Put the value in equation

2=A+0

A=2.....(I)

Now, on differentiating of general equation

x'(t)=-\sqrt{32}A\sin(\sqrt{32t})+\sqrt{32}B\cos(\sqrt{32t})

Using condition

(II). x'(0)=0

Then, x'(0)=-\sqrt{32}A\sin(\sqrt{32\times0})+\sqrt{32}B\cos(\sqrt{32\times0})

Put the value in the equation

0=0+\sqrt{32}B

So, B = 0

Now, put the value in general equation from equation (I) and (II)

So, The general solution is

x(t)=2\cos\sqrt{32t}

(e). We need to calculate the  time

Using formula of time

T=\dfrac{2\pi}{\omega}

Put the value into the formula

T=\dfrac{2\pi}{4\sqrt{2}}

T=1.11\ sec

Hence, (a). The value of spring constant is 4.

(b). The natural frequency is 4√2.

(c). The differential equation is \dfrac{d^2x}{dt^2}+32x=0

(d). The solution for the position is x(t)=2\cos\sqrt{32t}

(e). The time period is 1.11 sec.

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Answer:

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Explanation:

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8 0
2 years ago
A(n) 131 g ball is dropped from a height
larisa [96]

Answer:

26.59 N/m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 131 g

Extention (e) = 4.82755 cm

Acceleration due to gravity (g) = 9.8 m/s²

Spring constant (K) =?

Next, we shall convert 131 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

131 g = 131 g × 1 Kg / 1000 g

131 g = 0.131 Kg

Thus, 131 g is equivalent to 0.131 Kg.

Next, we shall the force exerted by the ball on the spring. This can be obtained as follow:

Mass (m) = 0.131 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = ma

F = 0.131 × 9.8

F = 1.2838 N

Next, we shall convert 4.82755 cm to metre (m)

This can be obtained as follow:

100 cm = 1 m

Therefore,

4.82755 cm = 4.82755 cm × 1 m / 100 cm

4.82755 cm = 0.0482755 m

Thus, 4.82755 cm is equivalent to 0.0482755 m

Finally, we shall determine the spring constant as follow:

Force (F) = 1.2838 N

Extention (e) = 0.0482755 m

Spring constant (K) =?

F = Ke

1.2838 = K × 0.0482755

Divide both side by 0.0482755

K = 1.2838 / 0.0482755

K = 26.59 N/m

Thus the spring constant is 26.59 N/m

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7 0
2 years ago
INSTI
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Answer:

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A 2100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact w
aliya0001 [1]

Answer:

   f = 878,080 N

Explanation:

mass of pile driver (m) = 2100 kg

distance of pile driver to steel beam (s) = 5 m

depth of steel driven (d) = 12 cm = 0.12 m

acceleration due to gravity (g0 = 9.8 m/s^{2}

calculate the average force exerted on the pile driver by the beam.

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              f x 0.12 = 2100 x 9.8 x (5- (-0.12))

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