Finding the period of small vertical oscillations

[green-beans]

Homework Statement

I need to find the period of small vertical oscillations about equilibrium position of a string whose motion can be described by the following equation:
d²x/dt² = (-g/h)*x

Answer: 2π√(h/g)

Homework Equations

I know that the time period is given by the formula
T = 2πω where ω = √({d²V/dx²} / m) where V is the potential

The Attempt at a Solution

I tried solving the differential equation but I got stuck by doing the following:
Transform differential equation into (by taking dx/dt on both sides):
d²x/dt² dx/dt = (-g/h)*x dx/dt
Denote dx/dt = m, then we can write:
∫m*dm/dt dt = ∫(-g/h)*x * dx/dt dt
(m)²/2 = -gx²/2h
which is:
(dx/dt)²/2 = -gx²/2h
which gives that dx/dt is equal to √(-gx²/h) which is impossible since there is a negative sign inside.

Thank you in advance!

 

[haruspex]

What about the constant of integration?

 

[lychette]

The equation of SHM is a = - ω² x
d²x/dt² Is the acceleration...can you see what is ω²
To find out what is t ?

T = 2Pi/ω ...not 2πω

 

[green-beans]

The equation of SHM is a = - ω² x
d²x/dt² Is the acceleration...can you see what is ω²
To find out what is t ?

T = 2Pi/ω ...not 2πω

Ohhh, thank you so much! So ω is √(g/h) and then T = 2π/ω = 2π√(h/g)
The thing is that we did not learn about this formula before and so I was wondering if there is a way to deduce T by solving the equation?

 

[lychette]

Ohhh, thank you so much! So ω is √(g/h) and then T = 2π/ω = 2π√(h/g)
The thing is that we did not learn about this formula before and so I was wondering if there is a way to deduce T by solving the equation?

The main thing you need to know is that acceleration is d²x/dt²
SHM is motion where acceleration is proportional to displacement...ie acc = - k x ...( - sign because restoring force, and therefore acceleration, is directed towards the equilibrium position)
In your case k = g/h

 

[haruspex]

I was wondering if there is a way to deduce T by solving the equation?

Yes, see post #2.

 

[green-beans]

Yes, see post #2.

Hi, thank you for your reply!
I totally forgot about the constant of integration. However, I still cannot get the needed result by just simply solving the equation.
So, with the constant of integration, I get:
dx/dt=v=√(c-gx²/h)
then I solve this differential equation again and get:
T = (√h) (arcsin ((x√g)/√ch) (in this case constant of integration is zero as at time t=0 the displacement from equilibrium position is zero)
But then I am not sure what I have to do with c.

 

[haruspex]

T = (√h) (arcsin ((x√g)/√ch)

That does not look right. There should be a g outside the arcsin.

 

[green-beans]

That does not look right. There should be a g outside the arcsin.

Ooops, I am really sorry - I made a slip. For the solution I got T = √(h/g) arcsin ({x√g}/{√ch}) but then I am not sure where does 2π in the solution come from :(

 

[haruspex]

Ooops, I am really sorry - I made a slip. For the solution I got T = √(h/g) arcsin ({x√g}/{√ch}) but then I am not sure where does 2π in the solution come from :(

It might be more obvious if you turn it around to express x as a function of time. What is the smallest increment of time after which the function repeats?

 

[green-beans]

It might be more obvious if you turn it around to express x as a function of time. What is the smallest increment of time after which the function repeats?

Ohhh, I see, thank you so much! It makes sense! :)