Question

Does \(sin^2\omega t\) represents SHM?

Answer 1

hmm haven't yet started SHM
still on thermodynamics
but yeah i know lil bit

okay well like i have seem a lil ques which are like this-

find the period of the function-
\(y=sin (\omega t) \) .. like this

are you talking about some term related to the topic which has that thing in it?

Answer 2

its just a simple function which represents the Simple Harmonic motion

Answer 3

See example 14.3
https://books.google.co.in/books?id=H0FiBgAAQBAJ&pg=PA341&lpg=PA341&dq=Which+of+the+following+functions+of+time+represent+(a)+simple+harmonic+motion+and+(b)+periodic+but+not+simple+harmonic?+Give+the+period+for+each+case.+(1)+sin+%CF%89+t+%E2%80%93+cos+%CF%89+t+(2)+sin2+%CF%89&source=bl&ots=u7o5oSldMt&sig=wOSay6G8UAtjM4tnlnTrW_Y1Es4&hl=en&sa=X&ved=0ahUKEwjUr7vl4P7JAhUSCo4KHWMCBAoQ6AEIKTAD#v=onepage&q=Which%20of%20the%20following%20functions%20of%20time%20represent%20(a)%20simple%20harmonic%20motion%20and%20(b)%20periodic%20but%20not%20simple%20harmonic%3F%20Give%20the%20period%20for%20each%20case.%20(1)%20sin%20%CF%89%20t%20%E2%80%93%20cos%20%CF%89%20t%20(2)%20sin2%20%CF%89&f=false

Answer 4

yeah in that case \(sin^2 \omega t\) is a function representing the harmonic motion

so we have - \(y=sin^2 \omega t\)

to become simple harmonic \(\large \frac{d^2y}{dt^2} ∝ y\)
here in our case
\(\large\frac{d^2y}{dt^2}=2 \omega^2[cos^2 \omega t-sin^2 \omega t]=2\omega^2\left( 1-\frac{2y}{a}\right)\)

yes it is directly proportional to y so its SHM

Answer 5

O_O

Answer 6

okay which part is troubling?

Answer 7

oh wait i did a lil mistake

Answer 8

Why have you taken -y and how do you know that for SHM double differentiation shoul dbe proportional to y

Answer 9

in calculation part

Answer 10

Yes.

Answer 11

i added that a from no where..

Answer 12

well.. i am not really sure about the reason :/
but well SHM's differential equation is this-
\(\large\frac{d^y}{dt^2}=\omega^2 y\)

so acc to this eq y must be directly proportion to the double derivative of the function for it to represent SHM

Answer 13

*\(\frac{d^2 y}{dt^2}\)

Answer 14

I am trying to understand how they have done in that book. Do you understand how they have done it?

Answer 15

It is harmonic motion with non-zero rest position and \(2\omega\) angular frequency.
\(sin^2(\omega t)=\frac12-\frac12\cos(2\omega t)=x_0+A\cos(\omega _0t+ \phi)\)

with

Rest position: \(x_0=\frac12\)
Amplitude: \(A=\frac12\)
Angular frequency: \(\omega_0 =2\omega\)
Phase at t=0: \(\phi=\pi\)

Answer 16

How \(sin^2(\omega t)=\frac12-\frac12\cos(2\omega t)\)

Answer 17

we know that
\(cos\theta =cos(\theta+2\pi)\)

so if we have our equation like this-> \(x=Acos(\omega t +\alpha)\)

then-
\(x=Acos(\omega t + \alpha + 2 \pi) =A cos \left( \omega\left( t+\frac{2 \pi}{\omega} \right)+\alpha\right)\)

lets say that \(t'=t+\frac{2\pi}{\omega}\)
so \(x=Acos(\omega t'+\alpha)\) the
this shows that the function at t coincides with the function at t'.
and the interval (t'-t) is the period of x

Answer 18

similarly in that link the equation was converted to this format so that we can find the time period easily

Answer 19

Humse naa ho payega bhaiya ye ab xD

Answer 20

lol xD

Answer 21

What *is* SHM? That's the real question.

Answer 22

Yes I think so..

Answer 23

I don't know that quite clearly in first place.

Answer 24

Q: What is SHM?
A: Yes I think so...

Full marks

Answer 25

Any function that can be expressed in the form of Asin(wt + \(\sf \phi\)). Correct?

Answer 26

That's not quite the definition of SHM, even though it is the solution to the equation.

Answer 27

Definitely not.

Answer 28

For instance:
\(\sin^2(\omega t)\) can, but \(\sin^3(\omega t)\) can't.

Answer 29

To check if anything is *SIMPLE* harmonic motion, check if its double-derivative is directly proportional to the position function. In other words, check if their ratio is constant. If it is, then yes.

Answer 30

My question for the English-speaking friends here is:
What does "simple" in simple harmonic motion refer to?
In French, we just have "Oscillateur harmonique", with no hint that there should be a simple type and a non-simple one.

Answer 31

\[x'' = -kx\]\[x = \sin^2 \omega t \]\[x' = 2\omega \sin \omega t \cos \omega t = \omega \sin (2\omega t) \]\[x'' = 2\omega^2 \cos (2\omega t)\]The required ratio is not constant.

@Vincent-Lyon.Fr In India, simple harmonic motion is introduced as any oscillation that can be modelled by the above equation.

Answer 32

Yes, but why not "Harmonic motion", full stop?

Answer 33

In that case your initial function does not represent a SHM since the rest-position is not zero.

Answer 34

Ah, I see what you're saying now. Nice.

Answer 35

Good... I imagine you're right.

Answer 36

However one wonders why the question is so ambiguous. There is no left-hand-side, i.e., saying something like \(x = \sin^2 \omega t\).

Answer 37

I find this case strange, because although it can be meaningful in maths to distinguish between \(A\sin(\omega _0t+ \phi)\) and \(A\sin(\omega _0t+ \phi)+x_0\), in physics a simple change in the choice of origin does not change the nature of the motion itself.

Answer 38

Well said. Didn't think of that one.

Answer 39

What I call simple harmonic motion, is a function \(x(t)\) that can be solution of a differential equation:

\(\ddot x(t)+\omega_0^2\;(x(t)-x_{rest})=0\)

but the definition might be different in India, imposing \(x_{rest}=0\) ;-)

Answer 40

It's obviously not different, lol. Just that in all contexts I've seen so far, the position function has been modeled in terms of the rest position, unless specified otherwise. I'll keep that one in mind though. Thank you very much.

Answer 41

Can any one explain me
How \(sin^2(\omega t)=\frac12-\frac12\cos(2\omega t)\)

Answer 42

\[\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = 1 - 2 \sin^2 \theta \]Now\[2 \sin^2 \theta = 1 - \cos (2 \theta)\]Divide both sides by two

Answer 43

@Abhisar what's the answer given to you?

Answer 44

Both following equations are very useful in physics:
\(2 \cos^2 \theta = 1 +\cos (2 \theta)\)
\(2 \sin^2 \theta = 1 - \cos (2 \theta)\)

Answer 45

Thanks, I got it now. I don't have a background in maths so i was puzzled how you did that.

Answer 46

Thank you very much for the time all of you guys :)

Answer 47

simple harmonic motion is motion pursuant to a **linear restorative** force. ie F = - kx.

so \(m \ddot x + k x = 0\)

with solution \(x(t) = A \cos (\omega t +\delta) \), same as \(x(t) = A \sin (\omega t +\delta) \)

i am pretty sure simple means no driver and no damping.

Answer 48

y=sin^2(wt) is an oscillatory function but does not represent shm. If you know about differentiation then differentiate y with respect to time t twice and you will find that you do not get a result which is a constant times the original function. For example if y=sin(at) then the second differential is -w^2 sin(wt) = constant times y. Your function does not produce such a result so it is not a representation of shm.

Answer 49

@vincent-lyon.fr
You have a good point about HM and SHM. Guess "oscillateur harmonique" covers more cases than SHM.
see also
https://en.wikipedia.org/wiki/Complex_harmonic_motion

Answer 50

@Vincent-Lyon.Fr , I think this is what you wanna trying to ask. Is it correct?

Answer 51

https://www.youtube.com/watch?v=SZ541Luq4nE

Answer 52

Awesome!

Answer 53

@Abhisar
IrishBoy123 gave the answer: 'simple' is opposed to 'driven' or 'damped'
It need not be linear.