Simple Harmonic Motion: Particle Speed and Acceleration Calculation

[Matt512]

Homework Statement

A particle moves with simple harmonic motion in a straight line with amplitude 0.05 m and the period 12 s.Find (a) the maximum speed , (b) the maximum acceleration of the particle.Write down the values of the constants P and Q in the equation
x/m = P sin [Q)(t/s)]

Homework Equations

The Attempt at a Solution

Well i tried to find the speed using A w sin w t
but my problem is , how do i find the t ?

 

[DrClaude]

$t$ stays a variable. You should be able to restrict your analysis to one oscillation, as the motion will be periodic.

 

[Matt512]

if i use the equation A w sin wt
i should replace t with what ?

 

[DrClaude]

You are asked to calculate the maximum speed and acceleration. Do you know how to do that? For that, you need to be able to write position as a function of time. You then need to find the times that will maximize each of these values, for which, as I hinted, you can restrict yourself to one oscillation.

By the way, the questions asks you to write it as x/m = P sin [Q(t/s)]. Use can use this instead of A and ω. Have you found P and Q?

 

[Matt512]

You are asked to calculate the maximum speed and acceleration. Do you know how to do that? For that, you need to be able to write position as a function of time. You then need to find the times that will maximize each of these values, for which, as I hinted, you can restrict yourself to one oscillation.

By the way, the questions asks you to write it as x/m = P sin [Q(t/s)]. Use can use this instead of A and ω. Have you found P and Q?

Its still not clear in my head :( Can you give me an example of how to write position as a function of time and where to take the max speed ?
I am still trying to do the part (a) , when i finish i will do the remaining parts

 

[DrClaude]

Can you give me an example of how to write position as a function of time

An example is given in the question:
$$
x = P \sin(Q t)
$$
You have position on the left-hand side, and a function of time on the right-hand side. Therefore, you can say that position is a function of time, that it varies with time.

I am still trying to do the part (a) , when i finish i will do the remaining parts

Finding P and Q is the first thing to do. I don't see how you can answer (a) or (b) without it. It may be the way the question is written that is confusing you. I read the last sentence of the problem as a hint, rather than a part (c) to be answered.

 

[Matt512]

Ah i see
Yeah the question was confusing me xD

Well
x=Psin(Qt) you compare it to
x=Asinwt

p will be equal to 0.05
and w(Q) will be π/6 right ?

 

[DrClaude]

Ah i see
Yeah the question was confusing me xD

Well
x=Psin(Qt) you compare it to
x=Asinwt

p will be equal to 0.05
and w(Q) will be π/6 right ?

Correct.

 

[Matt512]

Correct.

I have 1 question
When do you know that you have to use either A w cos wt or A w sin wt ?

 

[DrClaude]

I have 1 question
When do you know that you have to use either A w cos wt or A w sin wt ?

It depends on the initial conditions (and your choice of $t=0$). For a spring that is stretched at $t=0$ and then let go, the cosine will give you the right equation as you will get $x(0) = A$.

 

[DrClaude]

I should probably add that the general solution to the equation of motion of an harmonic oscillator is
$$
x(t) = A \cos (\omega t) + B \sin (\omega t)
$$
where $A$ and $B$ are set by the initial conditions (position and velocity at $t=0$). It is left as an exercise to show that this can be rewritten as
$$
x(t) = C \sin (\omega t + \phi)
$$

 

[Matt512]

Well if i choose X= A sin wt

then v=dx/dt = A w cos wt which is what i want

if i choose x = A cos wt it would be the inverse
right ?

 

[DrClaude]

Well if i choose X= A sin wt

then v=dx/dt = A w cos wt which is what i want

if i choose x = A cos wt it would be the inverse
right ?

Yes, and this corresponds, as I mentioned above, to the arbitrary choice of the zero of time.

 

[Matt512]

Well using v=dx/dt = A w cos wt i get 0.026 ms
What formulae should i use for the acceleration ? Is it the same principle as for the speed ?

 

[DrClaude]

Well using v=dx/dt = A w cos wt i get 0.026 ms
What formulae should i use for the acceleration ? Is it the same principle as for the speed ?

Yes, same principle.

 

[Matt512]

so its A w^2 cos wt ?

 

[DrClaude]

so its A w^2 cos wt ?

Yes, that's the equation for acceleration.

[Edit: but with a minus sign. Thanks to BvU for spotting this.]

 

[Matt512]

Thanks very much for you help :)

 

[BvU]

so its A w^2 cos wt ?

Actually, there should appear a minus sign when differentiating a cosine.
I hope that it is clear to you that when you start with

X= A sin wt

then v=dx/dt = A w cos wt

the acceleration is given by a = dv/dt = -A w² sin(wt)

And conversely, when you start with

X= A cos wt
then v=dx/dt = -A w sin(wt)
and a = dv/dt = -A w² cos(wt)​

 

[Matt512]

Actually, there should appear a minus sign when differentiating a cosine.
I hope that it is clear to you that when you start with the acceleration is given by a = dv/dt = -A w² sin(wt)

And conversely, when you start with

X= A cos wt
then v=dx/dt = -A w sin(wt)
and a = dv/dt = -A w² cos(wt)​

well for v and a i got the correct answer by using the cos equation
then when you derive the equation one should be cos and the other sin , for the v and a right?
and thd minus sign is it necessary to put??

 

[DrClaude]

well for v and a i got the correct answer by using the cos equation
then when you derive the equation one should be cos and the other sin , for the v and a right?
and thd minus sign is it necessary to put??

The minus sign is necessary to get the right time dependence for acceleration, $a(t)$. The problem asks for the maximum acceleration, which implicitly means the maximum in absolute value, so the answer will be positive.

 

[Matt512]

Thanks :)