Simple Harmonic Motion problem 1

[thunderhadron]

Hi friends the problem is -

https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-snc6/s480x480/6405_2656465868185_1414230035_n.jpg

Attempt -

As per the problem states,

For the first second equation of SHM, (using x = A sin ωt)

a = A sin ω

From here I get, sin ω = a/ A --------------(1)

Using it for the 2 seconds, total displacement would be a + b. So,

a + b = A sin 2ω

i.e. a + b = 2A sin ω. cos ω

i.e. a + b = 2A sin ω. √1-sin²ω ----------------(2)

Solving both the equations, I am getting the result,

A = 2a²/ √(3a² - b² - 2ab)

But the correct answer of this problem is option (A) as per the question.

Please friends help me in solving this Problem.

Thank you all in advance.

 

[TSny]

Note that the particle starts from rest. Does the equation x = Asinωt describe a particle starting from rest? (Assuming "start" means t = 0)

 

[thunderhadron]

Note that the particle starts from rest. Does the equation x = Asinωt describe a particle starting from rest? (Assuming "start" means t = 0)

No, thaks. I have taken the eqn in wrong manner.

Let the equation be x = A cosωt

Solving, a = A cos ω

i.e. cosω = a/A

And (a + b) = A cos 2ω

i.e. (a + b) = A √(2cos²ω -1)

Soving both we get A = √(a² - b² -2ab)

Yet answer is not achieved

 

[ehild]

The distance traveled can not be x=Acos(ωt), as the distance traveled would be A at zero time, (it should be zero) and the velocity would be negative...

If x(t) is the position at time t, the velocity is v=dx/dt. The particle travels in the positive x direction, and it is velocity is zero at t=0. What should be the form of the displacement as function of t?

ehild

 

[thunderhadron]

Creating such kind of function is really giving me problem. Please help me out.

 

[ehild]

What about Δx=A(1-cosωt)?

ehild

 

[thunderhadron]

What about Δx=A(1-cosωt)?

ehild

Yes this function is proper for the question. I appreciate your help. I will try to find answer with its help and get back to you soon.

 

[Saitama]

What about Δx=A(1-cosωt)?

ehild

I am currently doing SHM, I am willing to know how you arrived at that equation. :)

 

[thunderhadron]

I am currently doing SHM, I am willing to know how you arrived at that equation. :)

I think its answer.

In my notion he thought that -
We are not allowed to use x = A sin ωt or x = A cos ωt here because these functions are not giving the appropriate conditions according to question. But the function which can give correct answer for v = 0 at t = 0 is is x = A cos ωt. But it it is giving some displacement at t = 0.
so what it giving the displacement at t = 0. That is A. Subtract it from A and the displacement as well as the velocity both will be zero at t = 0.

Even I am not sure that the function is correct or not right now. I will solve it and then again will discuss about it.

 

[thunderhadron]

The distance traveled can not be x=Acos(ωt), as the distance traveled would be A at zero time, (it should be zero) and the velocity would be negative...
ehild

ehild,

Please tell me that why are saying that the displacement of the particle should be zero at t = 0 ?

If I stretch a spring mass system fro it mean position and hold it for some time and then release. So there the particle would have some displacement at t = 0. Isn't it ?

 

[thunderhadron]

What about Δx=A(1-cosωt)?

ehild

Thank you very much ehild I go the correct answer. But Please clear the doubt which raised earlier.

The function is absolutely correct.

 

[ehild]

Pranav: The shm along x in general is x=A cos (ωt+θ). The displacement from the position at zero time is Δx=x-x₀=Acos(ωt+θ)-x₀.

The particle starts from rest, so v=-Aωsinθ=0. That means θ=0 or π.
It starts to move in the positive x direction, v should increase, so the acceleration at t=0 is positive: -Aω²cosθ>0, that is θ=π.
cosθ=-1, but the displacement is zero if x₀=Acosθ=-A.
So you can write up the displacement as Δx=Acos(ωt+π)+A=A(1-cosωt).ehild

 

[ehild]

ehild,

Please tell me that why are saying that the displacement of the particle should be zero at t = 0 ?

If I stretch a spring mass system fro it mean position and hold it for some time and then release. So there the particle would have some displacement at t = 0. Isn't it ?

Displacement is the change of position Δx=x(t)-x(0). Of course, it is zero at t=0. It is not meant as the change of length of a spring here.

We start measuring time when the particle has zero velocity. Assuming a general SHM x(t)=Acos(ωt+θ), we can get theta from the conditions given, that the initial velocity is zero, and the particle has positive velocity after. Read my previous post.

ehild

 

[Saitama]

Pranav: The shm along x in general is x=A cos (ωt+θ). The displacement from the position at zero time is Δx=x-x₀=Acos(ωt+θ)-x₀.

The particle starts from rest, so v=-Aωsinθ=0. That means θ=0 or π.
It starts to move in the positive x direction, v should increase, so the acceleration at t=0 is positive: -Aω²cosθ>0, that is θ=π.
cosθ=-1, but the displacement is zero if x₀=Acosθ=-A.
So you can write up the displacement as Δx=Acos(ωt+π)+A=A(1-cosωt).


ehild

Thanks ehild!

 

[thunderhadron]

The shm along x in general is x=A cos (ωt+θ). The displacement from the position at zero time is Δx=x-x₀=Acos(ωt+θ)-x₀.

The particle starts from rest, so v=-Aωsinθ=0. That means θ=0 or π.
It starts to move in the positive x direction, v should increase, so the acceleration at t=0 is positive: -Aω²cosθ>0, that is θ=π.
cosθ=-1, but the displacement is zero if x₀=Acosθ=-A.
So you can write up the displacement as Δx=Acos(ωt+π)+A=A(1-cosωt).


ehild

Very nice explanation.