Displacment in simple harmonic motion after periodic time

In summary, the speaker is struggling to understand the concept of simple harmonic motion and how to calculate the periodic time using displacement. They mention using equations such as T=2*pi*sqrt(m/k) and T=2*pi*sqrt(m/f/x), but are confused about the value of x and whether it is equal to zero. They also mention their understanding of SHM, including acceleration, velocity, and amplitude, but are unsure of how to combine these laws to calculate the periodic time. They are seeking guidance on where to start and what to do.
  • #1
madah12
326
1
i just studied S.H.M and i am wondering is the displacement equals the amplitude or zero after the periodic time?
because when i thought it equals zero at first because the periodic time is time required for the object to make acomplete oscillation and return to it's starting point right?

but when i put the law T= [tex]2\pi\ \sqrt{m/k}[/tex]
T= [tex]2\pi\ *\sqrt{m/f/x}[/tex]

T= [tex]2\pi\ *\sqrt{m/mg/x}[/tex]

T=[tex]2\pi\ *\sqrt{x/g[/tex] (and unless the periodic time of every S.H.M is 0 this can't be right )and no this isn't home but i asked the tried to ask the teacher and he didnt have time to listen.
And this is like my first lesson in S.H.M so i don't understand it that much.
maybe i didnt understand the periodic time right or something like that

also since y = A Sin (wt)
y = Asin ([tex]2\pi/T\[/tex]*T
y = Asin([tex]2\pi[/tex]=0
 
Last edited:
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  • #2
Well if you start at max displacement then your equation for SHM will be of the form
y=Acos(wt)

if you start at zero minimum displacement, it will be of the form y=Asin(wt)
 
  • #3
But i still don't understand
if we try to calculate T
then we are trying to calculate the time the displacement reach 0 right?
so if the displacement is 0
then
T =[tex]\sqrt{Y/g}[/tex] right so if Y is zero wouldn't the T always be 0?(for the displacement i can use symbol X or Y right?)
 
  • #4
madah12 said:
T=[tex]2\pi\ *\sqrt{x/g[/tex] (and unless the periodic time of every S.H.M is 0 this can't be right )

What is x??
 
  • #5
displacement.
isn't it from hook's law
that f=x*k
so k= f/x
so i putted that value in the law for the periodic time
and i got T = root of x / root of g

Edit: i am really new to simple harmonic motion and i am not very good at physics therefore i am trying to understand the laws better but i still don't understand the concept of S.H.M
what i know of it is that its a motion of an object where the acceleration increase with the displacement but in the opposite direction and that oscillation is when the object goes through the max displacement to the right then to the left then return to where it started and the time for that is called the periodic time.

and that the object move with an angular speed (omega) which is the magnitude of the angular velocity which is measured by radian and it decrease when the acceleration increases.
and that the acceleration reach's it maximum when the object reaches the amplitude while the velocity become 0 and velocity reaches its maximum when the object return to it's starting place.(this is all i know about simple harmonic motion but i am trying to calculate T by displacement but always get zero as an answer)

these are the only laws i know
w=theta/t
w =[tex]2\pi\ /T[/tex]
y=Asin(wt)
T=[tex]2\pi\ *\sqrt{m/k}[/tex]
a=-w^2*Asin(wt)=w^2*y
amax=+or-w^2*A
V=w*[tex]\sqrt{A^2-y^2}[/tex]
Vmax=w*A
(fmax =k*A. i don't know if that is alaw since it isn't written in my book but it make since the returning force increase with the displacement it should be at max at the amplitude right?)
But i am trying to put alaw to combine the periodic time and the displacement but i can't do that.without always getting zeros

So i want to know what i am doing wrong anyone can help with that?
 
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  • #6
I don't really need some one to form the law for me i just want to make sure that the laws and info i have are correct and maybe little advice on where to begin
 

1. What is displacement in simple harmonic motion after one period?

The displacement in simple harmonic motion after one period is equal to the amplitude of the motion. This means that the object will return to its starting position after completing one full cycle.

2. How is displacement related to frequency in simple harmonic motion?

The displacement in simple harmonic motion is directly proportional to the frequency of the motion. This means that as the frequency increases, the displacement also increases.

3. Can displacement be negative in simple harmonic motion?

Yes, displacement can be negative in simple harmonic motion. This occurs when the object is moving in the opposite direction of its equilibrium position.

4. What is the formula for calculating displacement in simple harmonic motion?

The formula for calculating displacement in simple harmonic motion is: x = A*cos(2πft), where x is the displacement, A is the amplitude, f is the frequency, and t is the time.

5. Can the displacement in simple harmonic motion ever be greater than the amplitude?

No, the displacement in simple harmonic motion can never be greater than the amplitude. The amplitude represents the maximum displacement of the object, and the object will always return to this position after completing one full cycle.

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