Ask question

Ask question

All categories

2 years ago

7

The light elastic rod supports a 4 kg sphere. when an 18n vertical force is applied to the sphere, the rod deflects 14 mm. a) fi

nd the amplitude and natural frequency of the system when the 18n force is suddenly released. b) if there is an earthquake and the wall oscillates with a harmonic frequency of 2 hz and an amplitude of 15 mm up and down, determine the amplitude of vibration for the sphere.

1 answer:

Zinaida [17]2 years ago

8 0

The amplitude of vibration for the sphere will be 29.48 nm.

<h3 /><h3>What is amplitude?</h3>

The distance between from the crest of a wave to the next crest is known as the amplitude.

The amplitude of the sphere is;

$\rm A= \frac{\delta_0}{[1-(\frac{\omega}{\omega_n}^2]} \\\\$

The stiffness of the load is;

$\rm K = \frac{F}{\triangle y }\\\\ K=\frac{18}{0.014} \\\\ K=1285.71 \ N/m$

The natural frequency is;

$\rm \omega_n= \sqrt{\frac{k}{m}} \\\\ \rm \omega_n= \sqrt{\frac{1285.71}{4}} \\\\ \omega_n=17.93 \ rad/sec$

Put the obtained values in the amplitude formula;

$\rm A= \frac{0.015}{[1-(\frac{4 \pi}{17.93}^2]} \\\\ A= 0.02948 \\\\ A= 29.48 \ nm$

Hence,the amplitude of vibration for the sphere will be 29.48 nm.

To learn more about the amplitude, refer:

brainly.com/question/8662436

#SPJ1

You might be interested in

Say you want to make a sling by swinging a mass M of 1.9 kg in a horizontal circle of radius 0.042 m, using a string of length 0

padilas [110]

Answer: T= 715 N

Explanation:

The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r

At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.

So the kinetic energy will be the following:

K = 1/2 m v² = 15. 0 J

Solving for v², and replacing in the expression for T:

T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N

3 0

3 years ago

One complete expression of a waveform beginning at a certain point, progressing through the zero line to the wave’s highest (cre

elixir [45]

Answer:

wavelength.

Explanation:

One complete expression of a waveform beginning at a certain point, progressing through the zero line to the wave’s highest (crest) and lowest (trough) points, and returning to the same value as the starting point is called a is called wavelength. Its can be also defined as the distance between two successive crests or trough points in wave form.

4 0

3 years ago

A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en

Brut [27]

Answer:

75 N

Explanation:

In this problem, the position of the crate at time t is given by

$y(t)=2.80t+0.61t^3$

The velocity of the crate vs time is given by the derivative of the position, so it is:

$v(t)=y'(t)=\frac{d}{dt}(2.80t+0.61t^3)=2.80+1.83t^2$

Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:

$a(t)=v'(t)=\frac{d}{dt}(2.80+1.83t^2)=3.66t$ [m/s^2]

According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:

$F(t)=ma(t)$

where

m = 5.00 kg is the mass of the crate

At t = 4.10 s, the acceleration of the crate is

$a(4.10)=3.66\cdot 4.10 =15.0 m/s^2$

And therefore, the force on the crate is:

$F=ma=(5.00)(15.0)=75 N$

7 0

3 years ago

The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 0 › m/s in 0.02 s, due to the gravitational attraction of

Alenkasestr [34]

Answer:

a) $a=(-2,-6,0)m/s^2$ , with a magnitude of $6.3m/s^2$

b) $\frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2$ , with a magnitude of $0.76Kgm/s^2$

c) $F=(-0.24,-0.72,0)N$ , with a magnitude of $0.76N$

Explanation:

We have:

$v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg$

We can calculate each component of the acceleration using its definition $a=\frac{\Delta v}{\Delta t}$

$a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\$

The rate of change of momentum of the ball is $\frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma$

So for each coordinate:

$\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2$

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

$a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N$

<em>(N and </em> $Kgm/s^2$ <em> are the same unit).</em>

3 0

3 years ago

When mantle rocks near the radioactive core are heated, they become less dense than the cooler, upper mantle rocks. These warmer

avanturin [10]

C.convection cells is the right answer

4 0

3 years ago

Read 2 more answers