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3 years ago

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A force of 9 pounds stretches a spring 1 foot. A mass weighing 6.4 pounds is attached to the spring, and the system is then imme

rsed in a medium that offers a damping force numerically equal to 1.2 times the instantaneous velocity. Find the equation of motion If the mass is initially released from rest from a point 1 foot above the equilibrium position.

1 answer:

mart [117]3 years ago

7 0

Answer:

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

Explanation:

let $m$ be the mass attached, let $k$ be the spring constant and let $\beta$ be the positive damping constant.

-By Newton's second law:

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

where $x(t)$ is the displacement from equilibrium position. The equation can be transformed into:

$\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0$ shich is the equation of motion.

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(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa

Vedmedyk [2.9K]

Missing data: the wave number

$k=60 cm^{-1}$

(a) $z = 0.003 sin (6000y-31.4 t)$

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

$z = A sin (ky-\omega t)$

where

A is the amplitude of the wave

k is the wave number

$\omega$ is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

$k = 60 cm^{-1} = 6000 m^{-1}$ is the wave number

$T = 0.20 s$ is the period, so the angular frequency is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s$

So, the wave equation (in meters) is

$z = 0.003 sin (6000y-31.4 t)$

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

$v_t = z' = -A \omega cos (ky-\omega t)$

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

$v_{t}_{max} =\omega A$

where

$\omega = 31.4 rad/s\\A = 0.003 m$

Substituting,

$v_{t}_{max}=(31.4)(0.003)=0.094 m/s$

(c) 5.24 mm/s

The wave speed is given by

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

The frequency can be found from the angular frequency:

$f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz$

While the wavelength can be found from the wave number:

$\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m$

Therefore, the wave speed is

$v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s$

7 0

3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

2 years ago

A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p

Andrei [34K]

Answer:

13.4 x 10 raise to power -19 C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

. The uniform electric field is E = 214 N/M

, The decrease in electrical potential energy is d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

which is given by the equation

d(P.E) =qEd

51.63 x 10 power -19 = q(214)(0.018)

51.63 x 10 power -19 =3.852q

by making q the formular,

q = 13.4 x 10 power -19 C

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3 years ago

What’s a good hypothesis and and experiment for water?

Lelu [443]

Answer: You could do something like, "how does water react to being mixed with baking soda"...or something along those lines

Explanation:

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3 years ago

A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1

pav-90 [236]

I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
$F_a=F_cf-F_g$
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
$F_a=m\frac{v^2}{r}-mg=179 $N$
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At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
$F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N$
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
$F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}$

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3 years ago