# Homework Help: Finding the period of small vertical oscillations

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1. Nov 5, 2016

### green-beans

1. The problem statement, all variables and given/known data
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

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

3. 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.

2. Nov 5, 2016

### haruspex

What about the constant of integration?

3. Nov 5, 2016

### 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

4. Nov 5, 2016

### green-beans

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?

5. Nov 5, 2016

### lychette

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

6. Nov 5, 2016

### haruspex

Yes, see post #2.

7. Nov 7, 2016

### green-beans

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.

8. Nov 7, 2016

### haruspex

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

9. Nov 7, 2016

### green-beans

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 :(

10. Nov 7, 2016

### haruspex

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?

11. Nov 8, 2016

### green-beans

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