Simple harmonic motion -- manipulating SHM equations

[jisbon]

Homework Statement

Object doing S.H.M , displacement $D(t)=Asin(\omega t)$
where A is amplitude, $\omega$ is the angular frequency. Speed of object is v when displacement is half of the amplitude. Find speed in terms of v when displacement is 0

Relevant Equations

NIL

My workings:
$D(t) = Asin\omega t$
$v(t) = \frac{\text{dD}}{\text{dt}}=Acos(\omega t)\omega$
$v(t) =Acos(\omega t)\omega$
When displacement half of amplitude,
$Asin\omega t$ = 0.5$A$
$sin\omega t$ = 0.5
$v(t) =Acos(\omega t)\omega$
$v(t) =\omega (0.5Asin\omega t)cos \omega t$
$v = \omega (0.5A*0.5)cos \omega t$
Pretty confused here on what I should do to proceed on. Any ideas?

 

[Orodruin]

What is cos(q) when sin(q) = 0.5?

 

[jisbon]

What is cos(q) when sin(q) = 0.5?

$\sqrt{\frac{3}{4}}$

So, I can sub
$v = \omega (0.5A*0.5)cos \omega t$
$v = \omega (0.5A*0.5) (\sqrt{\frac{3}{4}})$
? Do I simplify it?

 

[haruspex]

$\sqrt{\frac{3}{4}}$

So, I can sub
$v = \omega (0.5A*0.5)cos \omega t$
$v = \omega (0.5A*0.5) (\sqrt{\frac{3}{4}})$
? Do I simplify it?

It says: Find speed in terms of v

 

[jisbon]

It says: Find speed in terms of v

Simplifying this will make $v= \frac{\sqrt{3}}{8}\omega A$
So this is when displacement half of amplitude.
So when displacement = 0, how do I relate this back to the equation above?

 

[Orodruin]

Your expression with the parentheses is wrong.

 

[Delta2]

your equation $v(t)=A\omega\cos{\omega t}$ is correct.
When displacement is zero what is $\cos{\omega t}$? What is the velocity then?

 

[jisbon]

When displacement is 0,
$Asin \omega t = 0$?

Your expression with the parentheses is wrong.

As in v(t) or?

 

[Orodruin]

As in what happened in this step is incorrect

$v(t) =\omega (0.5Asin\omega t)cos \omega t$

 

[jisbon]

As in what happened in this step is incorrect

Reason I got this is because $v= \omega Acos\omega t$, and $A = 0.5A sin \omega t$, since they state that the displacement its half of its amplitude, when displacement is given as $Asin\omega t$?

 

[Delta2]

Reason I got this is because $v= \omega Acos\omega t$, and $A = 0.5A sin \omega t$, since they state that the displacement its half of its amplitude, when displacement is given as $Asin\omega t$?

That equation $A = 0.5A sin \omega t$ is not correct (you get after simplification of A that $1=0.5\sin{\omega t}\Rightarrow \sin{\omega t}=2$ which simply cannot hold.

 

[jisbon]

Al

That equation $A = 0.5A sin \omega t$ is not correct (you get after simplification of A that $1=0.5\sin{\omega t}\Rightarrow \sin{\omega t}=2$ which simply cannot hold.

Alright I understand that the equation is now impossible to achieve. So right now the only equation I can only work with is $v= \omega Acos \omega t$?

 

[Delta2]

Al

Alright I understand that the equation is now impossible to achieve. So right now the only equation I can only work with is $v= \omega Acos \omega t$?

yes that is a correct equation. Also from post #8 when displacement is zero then $A\sin{\omega t}=0\Rightarrow \sin{\omega t}=0$. What is $\cos{\omega t}$ then. Replace it in the above correct equation . And you find the speed (when displacement is zero) as a function of A and $\omega$. Now all you got to do is use the other condition (when displacement is half of A the speed is V) to express A and $\omega$ as function of V.

 

[Orodruin]

Al

Alright I understand that the equation is now impossible to achieve. So right now the only equation I can only work with is $v= \omega Acos \omega t$?

As has already been said, there is nothing wrong woth that equation. However, let me try a bit of genersl advice. In both this thread and your thread on arc length, you have gone from expressions that are relatively easy to work with to different expressions that are not at all as straightforward for giving you the answer. In both cases this has led to getting you confused and/or the wrong result. Therefore I would suggest spending some time trying to get better at identifying your target in problems as it feels you are to some extent rushing in applying formulae without first thinking about where they are going to get you.

 

[jisbon]

yes that is a correct equation. Also from post #8 when displacement is zero then $A\sin{\omega t}=0\Rightarrow \sin{\omega t}=0$. What is $\cos{\omega t}$ then. Replace it in the above correct equation . And you find the speed (when displacement is zero) as a function of A and $\omega$. Now all you got to do is use the other condition (when displacement is half of A the speed is V) to express A and $\omega$ as function of V.

Alright so from what you stated, I managed to find out that $cos \omega t =1$
Hence $v=\omega A$
So this is the velocity of v when displacement is 0

On the other hand when displacement is half of the amplitude, I will need to find the velocity, which equates to $v =\omega Acos \omega t$
From here, what I see is I only can proceed by subbing A with $Asin\omega t /2$ Is this still correct?

 

[Orodruin]

You are now confusing yourself by using the same notation ($v$) for the velocity at half the amplitude and the sought velocity.

 

[jisbon]

You are now confusing yourself by using the same notation ($v$) for the velocity at half the amplitude and the sought velocity.

Doesn't $v =\omega Acos \omega t$ apply for all ampltidues?

 

[Orodruin]

Doesn't $v =\omega Acos \omega t$ apply for all ampltidues?

No, because $v$ is here given as the speed at half the amplitude in the problem formulation. If you replace $v$ by something denoting the velocity at a particular time, call it $u$, then it is correct.

 

[jisbon]

No, because $v$ is here given as the speed at half the amplitude in the problem formulation. If you replace $v$ by something denoting the velocity at a particular time, call it $u$, then it is correct.

OK so let me assume v is the velocity when the displacement is half of the amplitude and u velocity when displacement is zero.
So from the previous equation that I managed to derive, $u = \omega A$
Now to derive v, will v now be as follows:
$v= \omega Acos\omega t$
$v= \omega (0.5Asin\omega t) cos \omega t$?

 

[Delta2]

Doesn't $v =\omega Acos \omega t$ apply for all ampltidues?

Since it is a function of time t, it would be more proper to say that is holds for all times t.

So let's call $t_1$ the time that the displacement is half of the amplitude and $t_2$ the time that the displacement is zero.
Then by using the above correct equation that holds for all times t , hence applying it for $t=t_1$ we get
$v(t_1)=A\omega\cos\omega t_1$. But it is given that at time $t_1$ it is $v(t_1)=v$, so after all we have the equation
$v=A\omega\cos\omega t_1$. Also because at time $t_1$ the displacement is half of the amplitude we get the equation $A/2=A\sin\omega t_1$ hence we can find that $\sin\omega t_1=1/2$ and hence $\cos\omega t_1=\sqrt\frac{3}{4}$.
So after all we come up with the equation $v=A\omega\sqrt\frac{3}{4}$

Now work in a similar way for the time $t=t_2$ when the displacement is zero. What equation will you come up with?

 

[Orodruin]

Since it is a function of time t, it would be more proper to say that is holds for all times t.

No it would not. The v given in the problem does not depend on time so the expression is wrong unless you specify that the time is that when the displacement is half the amplitude.

Now to derive v, will v now be as follows:
v=ωAcosωtv=ωAcosωtv= \omega Acos\omega t
v=ω(0.5Asinωt)cosωtv=ω(0.5Asinωt)cosωtv= \omega (0.5Asin\omega t) cos \omega t?

First, you do not derive v, it is given in the problem. Second, you are still making the same mistake that has been pointed out to you.

 

[jisbon]

No it would not. The v given in the problem does not depend on time so the expression is wrong unless you specify that the time is that when the displacement is half the amplitude.First, you do not derive v, it is given in the problem. Second, you are still making the same mistake that has been pointed out to you.

The only v that is present is the v that I have derived by differentiating the displacement. How do I proceed from the same equation?

 

[Delta2]

No it would not. The v given in the problem does not depend on time so the expression is wrong unless you specify that the time is that when the displacement is half the amplitude.

Yes, sorry I had in my mind the equation $v(t)=A\omega\cos\omega t$ which certainly holds for all times t. If we take $v$ to be the velocity at the time instance $t_1$ when displacement is half of amplitude then what it holds is $v=A\omega\cos\omega t_1$.

 

[Orodruin]

I would use a different notation than $v(t)$ to avoid any confusion.

 

[jisbon]

Thanks all who reply :) Managed to solve this with @Delta2 explanation actually Answer is $\frac {2}{\sqrt 3}v$

 

[PeroK]

Thanks all who reply :) Managed to solve this with @Delta2 explanation actually Answer is $\frac {2}{\sqrt 3}v$

I don't see how you can work a problem like this without labelling some specific times. After looking at the equations, the first thing I would do is:

Let $t_0, t_1$ be times when the displacement is zero and half the amplitude respectively.

Then you have $D(t_0) = 0 \ \Rightarrow \ \sin(wt_0) = 0$

Also, strictly speaking, you should be dealing with the absolute value for the speed. Alternatively, you could make a simplifying assumption that the displacements and velocities are positive, hence the sines and cosines are positive, at $t_0, t_1$.