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Sergio039 [100]
3 years ago
12

Simple Harmonic Motion

Physics
1 answer:
Delvig [45]3 years ago
7 0

- The net force is greatest at the position of maximum displacement

- The net force is zero when at the equilibrium position

Explanation:

The motion of a spring is a Simple Harmonic Motion, in which the displacement of the end of the spring is given by a periodic function of the form

x=Asin (\omega t)

where A is the amplitude (the maximum displacement), and \omega the angular frequency of the motion.

We can analyze the net force acting on the spring by looking at Hooke's law:

F=kx

where

F is the net force

k is the spring constant

x is the displacement

From the equation, we notice immediately that:

  • The net force is the greatest when the displacement x is the greates, so at the position in which the spring has maximum compression or stretching
  • The net force is zero when the displacement x is zero, so when the spring crosses the equilibrium position

Learn more about forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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A

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Two astronauts, each having a mass of 74.3 kg are connected by a 13.1 m rope of negligible mass. They are isolated in space, orb
murzikaleks [220]

Answer:

  L = 5076.5 kg m² / s

Explanation:

The angular momentum of a particle is given by

         L = r xp

         L = r m v sin θ

the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1

as we have two bodies

       L = 2 r m v

let's find the distance from the center of mass, let's place a reference frame on one of the masses

        x_{cm} = \frac{1}{M} \sum  x_{i} m_{i}i

        x_{cm} = \frac{1}{m+m} ( 0 + l m)

        x_{cm} = \frac{1}{2m}  lm

        x_{cm} = \frac{1}{2}

        x_{cm} = 13.1 / 2 = 6.05 m

let's calculate

          L = 2  6.05  74.3  5.65

          L = 5076.5 kg m² / s

4 0
3 years ago
Read 2 more answers
Can someone help me with universal gravation ​
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States that particles are attracts with every other particle. wich force is directily proportional product of two masses and inversely proportional to the distance between the centers.
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3 years ago
A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o
Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m

The fundamental frequency of the wire is given by

f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N

The tension in the wire is 67.9064 N

7 0
3 years ago
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