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Serga [27]
2 years ago
10

Suppose you have a pendulum clock that keeps correct time on Earth (acceleration due to

Physics
2 answers:
vekshin12 years ago
8 0

Answer:

dont know because I am a student lol

Alenkinab [10]2 years ago
7 0
9.8 minutes approximately

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As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horiz
Eddi Din [679]

a)E= U + K = \frac{1}{2}kx² +  \frac{1}{2}mv²

The total energy of the system at any point in the motion is equal to the sum of the elastic potential energy of the spring, U, and of the kinetic energy of the mass, K:

E= U + K = \frac{1}{2}kx² +  \frac{1}{2}mv²

where

'k' represents the spring constant

'x' is the compression/stretching of the spring with respect to its equilibrium position

'm' is the mass of the block attached to the spring

and 'v' is the speed of the block

b) <em>A=</em>\sqrt{\frac{2E}{k}}<em> </em>

The amplitude of the motion compares to the most extreme displacement of the mass-spring system. The displacement of the system, x(t), at time t, for a simple harmonic oscillator is given by,

x= Asin(ωt+∅)

where

amplitude  is 'A'

\omega=\sqrt{\frac{k}{m}} is the angular frequency of the motion

t is the time

\phi is the phase (we can take \phi=0 )

The amplitude of the motion occurs when the displacement of the motion is maximum: x=A. Regarding energy, the mass-spring system is at its maximum displacement (x=A) when all the mechanical energy of the framework is elastic potential energy, so when the kinetic energy is zero:

K=\frac{1}{2}mv^2=0

E=\frac{1}{2}kA^2\\ -->(1)

<em>A=</em>\sqrt{\frac{2E}{k}}<em> </em>

c)v_{max}=\omega A<u></u>

When the elastic potential energy is zero, the maximum speed of the system occurs i.e U=0 and the kinetic energy is maximum, so:

U=0

E=\frac{1}{2}mv_{max}^2

According to the law of conservation of the mechanical energy, this energy must be equal to the energy of the system at its maximum displacement (1), so we can write

\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2

and solving for v_{max}we find an expression for the maximum speed:

v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{k}{m}}A=\omega A

<h2><u></u>v_{max}=\omega A<u></u></h2>
4 0
3 years ago
Define Momentum? ......
puteri [66]
An object's momentum is defined as the product of its mass and speed.
6 0
3 years ago
Read 2 more answers
You need to determine the density of a ceramic statue. if you suspend it from a spring scale, the scale reads 32.4 n . if you th
Mandarinka [93]
Well, the density of the water is
{1000 \frac{kg}{ {m}^{3} } }
so i believe that is what the question is asking for :)
4 0
3 years ago
What happened to the maximum height of consecutive swings
GarryVolchara [31]

Answer:

we need more info

Explanation:

3 0
3 years ago
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
Lemur [1.5K]

Answer:

a)   v = 0.9167 m / s,  b)  A = 0.350 m,  c)  v = 0.9167 m / s, d)  A = 0.250 m

Explanation:

a) to find the velocity of the wave let us use the relation

          v = λ f

the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength

           λ = x

           λ = x

the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period

          T / 2 = t

           T = 2t

period and frequency are related

           f = 1 / T

           f = 1 / 2t

we substitute

           v = x / 2t

           v = 5.50 / 2 3

           v = 0.9167 m / s

b) the amplitude is the distance from a maximum to zero

          2A = y

           A = y / 2

           A = 0.700 / 2

           A = 0.350 m

c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same

      v = 0.9167 m / s

d) the amplitude is

           A = 0.500 / 2

           A = 0.250 m

4 0
3 years ago
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