Free Oscillations of materials

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The discussion revolves around a vertical oscillating system with three equal springs and a mass of 2 kg, where the motion is described by the equation x(t) = 0.4 cos(4t) + 0.5 sin(4t). Participants seek to determine the equivalent elasticity coefficient, initial conditions, natural frequency, oscillation period, and amplitude. Key points include the relationship between displacement and velocity, with the velocity at t=0 calculated to be 2. The conversation emphasizes understanding derivatives and the connection between oscillation parameters, ultimately aiding high school students in grasping these concepts.
adu1905
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Problem

A vertical oscillating system consist of three equal springs with elasticity coefficient c , among which a mass m=2kg is suspended.The system oscillates according to the following law of motion x(t)=0,4 cos 4t+0,5 sin 4t , (m) .

Determine:

1) The equivalent elasticity coefficient ce of reduced spring.
2) For t=0 values for initial conditions x0 and v0
3) The natural frequency k
4) The oscillation period T
5) The amplitude A


Note = Please some help. I am high school student .Give me tips then i will try. Thanks to who helps.
 

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I got some ideas for the 3rd part of question.
I have to find natural frequency. And for this need The natural period of the oscillation .
And the natural period of the oscillation formulas in the attachment 2. (down)
And the naturel frequency formula is it the attachment 1. (up)
This are true or not ? And natural period of the oscillation means oscillation period ?
 

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1) The force by a spring is given as F = -kx, where x is the displacement and k is the elasticity coefficient, if you move your mass a small amount what is the net force on it from your three springs?
2) You just need to evaluate x(0) and v(0), if you're stuck with the second see if you can remember how v is linked to x.
3) There is a formula for the frequency of a spring given the elasticity coefficient and the mass.
4) Period and frequency are very closely related, see if you can find the formula to convert them.
5) The mass will be at its furthest points when v = 0.
 
The equation of a simple harmonic motion is
x(t) =Asin(ωt+θ),
where ω is the angular frequency. It is 2pi times the natural frequency f :
ω = 2p*f. The time period is the reciprocal of the frequency: T=1/f.
A is the amplitude of the SHM and θ is a phase constant. x(t) is equivalent to the sum of a pair of sine and cosine functions of the same ωt:

x(t) =Asin(ωt+θ)=(Acos(θ))sin(ωt)+(Asin(θ))cos(ωt)=asin(ωt)+bcos(ωt).

The amplitude A is related to the coefficients of the sine and cosine terms on the right-hand side: A2=a2+b2.

Compare the function in the problem with the function in the equation above. What is the frequency and period? What is the amplitude A of the equivalent single SHM?

ehild
 
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Can you help whit numbers because my english very bad.Its not my main language :(
x(t) =asin(ωt)+bsin(ωt) this is formul. And
x(t)=0,4cos4t+0,5sin4t how can i compare .
And note i am in high school.I thınk this kind of questions are hard for us ...I don't know where can i start.Please create equations whit my number .Thank you very much .
 
adu1905 said:
Can you help whit numbers because my english very bad.Its not my main language :(
x(t) =asin(ωt)+bsin(ωt) this is formul. And
x(t)=0,4cos4t+0,5sin4t how can i compare .
Sorry, I meant x(t) =asin(ωt)+bcos(ωt).
Your equation is x(t)=0.5sin(4t)+0.4cos(4t)

a sin(ωt) ≡ 0.5 sin (4t)

b cos(ωt) ≡ 0.4 cos(4t)

Numbers and symbols of the same colour are identical.



ehild
 
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I think i found some answers.in the attachments...Please can you check ? this are 3-4-5 part of my question.What about the first and second part.

For second part. (For t=0 values for initial conditions x0 and v0 )

x(0) = 0,4 cos (4x0)+0,5 sin (4x0) ===> Cos 0 = 1 and Sin 0 = 0
x(0) = 0,4

is this true ? and what does v0 means ?
 

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Your answers are correct.
v0 is the velocity at time t=0. The velocity is the derivative of the displacement:
Do you know what derivative is?
ehild
 
Yes i know a little.First derivative of displacement is velocity.And second derivative of displacement is acceleration.
So İf t=0 then x0=0,4 ...And first derivative of x(t)=0,4cos4t+0,5sin4 is 0 isn't it ?
Then Vo is equal 0 ? ( Derivative of constant number is 0 ) ? :S
 
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  • #10
First do the derivation, then substitute t=0. Do you know the derivatives of cosine and sine?

If derivation is not clear to you yet, you certainly was shown that the oscillatory motion can be imagined as projection of uniform circular motion. See picture. A point moves along a circle of radius R with angular velocity ω. At any angular position α, x=Rcos(α) and y=Rsin(α). The velocity V is tangent to the circle, and its magnitude is ωR. The components of the velocity vector are Vx=-Rωsin(α), Vy=Rωcos(α).

It is a uniform circular motion, α=ωt. You can imagine that a function f(t)= Acos(ωt) is the x component of a circular motion along a circle of radius R=A, and the function g(t)=Asin(ωt) is the y component of the same circular motion. That is true also for the velocities, which are equal to the time derivatives of f and g.
f'=Vf = -Aωsin(ωt) and g'=Vg=Aωcos(ωt).

Your function is the sum of one sine and a cosine. The derivatives add up. Can you proceed? ehild
 

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  • #11
@ehild I am very happy to meet whit you.Thank you very much.You helped me and tried to teach my classmates.We will get higher marks i am sure .Thanks again .
 
  • #12
You are welcome. Hoping to meet you here soon again.:smile:

ehild
 
  • #13
Hi again.My teacher told me everythink is true except elasticity coefficient ce of reduced spring - and Vo for t=0, if the law of motion x(t) is given x(t)=0,4 cos 4t+0,5 sin 4t ...Thanks again.
 
  • #14
adu1905 said:
Hi again.My teacher told me everythink is true except elasticity coefficient ce of reduced spring - and Vo for t=0, if the law of motion x(t) is given x(t)=0,4 cos 4t+0,5 sin 4t ...Thanks again.

If x(t) = 0.4 cos (4t)+0.5 sin(4t) the velocity is -0.4*4 sin(4t) +0.5*4 cos(4t).
At t=0, v(0) = 2.

ehild
 
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