Helium balloon and simple harmonic motion

[~christina~]

Homework Statement

A light balloon filled wiht helium and with a density of 0.180kg/m^3 is tied to a light string of length L= 3.00m. The string is tied to the ground, forming an "inverted" simple pendulum. The balloon is displaced slightly from it's equillibrium position, as shown in picture.
a) show that the balloon executes simple harmonic motion when it is released, i.e., show that the restoring force is a Hooke's law force.

b) What is the period of the motion?

Ignore energy lost due to air friction and take the density of air to be 1.29kg/m^3
http://img212.imageshack.us/img212/9724/58435914sp1.th.jpg

Homework Equations

?

The Attempt at a Solution

a) show that the balloon executes simple harmonic motion when it is released, i.e., show that the restoring force is a Hooke's law force.

not sure how to prove this and how it should look if I do prove it when I do.
in the beginning position a):

balloon has:
buoyant force: up
gravity: down
tension: down

position b:
balloon
buoyant force: up
gravity: down?
tension: at an angle

forces on balloon
$$\Sum F= B-mg -T= 0$$

$$\Sum F'= B-mg- Tsin \theta$$

I really don't know... do I combine them??

b) period of the motion

don't know how to find this as well...

please help me on this.

Thank you

 

[kamerling]

(B-mg) is a constant force directed upwards. This must be balanced by the vertical component of the tension in the rope: T cos(theta). The horizontal component of the
tension in the rope is your restoring force.

For small values of theta you can replace cos(theta) by 1 and sin(theta) by theta
(if theta is in radians).

 

[~christina~]

(B-mg) is a constant force directed upwards. This must be balanced by the vertical component of the tension in the rope: T cos(theta). The horizontal component of the
tension in the rope is your restoring force.

For small values of theta you can replace cos(theta) by 1 and sin(theta) by theta
(if theta is in radians).

Um isn't it supposed to be Tsin(theta) for the restoring force?

so it would be

$$\Sum F= B-mg -T= 0$$

$$\Sum F'= B-mg- Tsin \theta$$

subtract them

$$-B+mg+T= 0$$
$$B-mg-Tsin \theta$$
_______________________
$$F_{net}= T-Tsin \theta$$

I'm not sure how do I know it's simple harmonic motion.

and how do I find part b)
The period of the motion? using the density they gave?

 

[rock.freak667]

The restoring force is $Tsin\theta$ and if $theta$ is small and in radians then $sin\theta \approx \theta$

So that in the horizontal direction.

$$ma= -T\theta$$and the distance it is displaced is given by the length of the arc (x) multiplied by the length,L, i.e. $\theta=xL$

 

[~christina~]

The restoring force is $Tsin\theta$ and if $theta$ is small and in radians then $sin\theta \approx \theta$

So that in the horizontal direction.

$$ma= -T\theta$$


and the distance it is displaced is given by the length of the arc (x) multiplied by the length,L, i.e. $\theta=xL$

but how would that help in my proving that the balloon undergoes simple harmonic motion?

was what I did above incorrect?

 

[rock.freak667]

but how would that help in my proving that the balloon undergoes simple harmonic motion?

was what I did above incorrect?

Well to show that something undergoes SHM, you need to show that $a=-\omega^2x$ where x=displacement,a=acceleration.

 

[~christina~]

ok..I'm assuming that you were going off what I did before so:

$$\Sum F= B-mg -T= 0$$

$$\Sum F'= B-mg- Tsin \theta$$

subtract them

$$-B+mg+T= 0$$
$$B-mg-Tsin \theta$$
_______________________
$$F_{net}= T-Tsin \theta$$

and including what you said where $$sin\theta\aprox \theta$$ then

$$F_{net}= T-Tsin \theta$$

I know that F= ma however you showed that

The restoring force is $Tsin\theta$ and if $theta$ is small and in radians then $sin\theta\approx \theta$

So that in the horizontal direction.

$$ma= -T\theta$$

and the distance it is displaced is given by the length of the arc (x) multiplied by the length,L, i.e. $\theta=xL$

why is it not $$ma= T-T\theta$$ ?

Well to show that something undergoes SHM, you need to show that $a=-\omega^2x$ where x=displacement,a=acceleration.

 

[rock.freak667]

Because the restoring force is given by Tsin. When you split T into its components you must use the components and not the Resultant vector T

 

[~christina~]

Because the restoring force is given by Tsin. When you split T into its components you must use the components and not the Resultant vector T

ok.. so it cancels out...

I'm assuming that you were going off what I did before so:

$$\Sum F= B-mg -T= 0$$

$$\Sum F'= B-mg- Tsin \theta$$

subtract them

$$-B+mg+T= 0$$
$$B-mg-Tsin \theta$$
_______________________
$$F_{net}= T-Tsin \theta$$

and including what you said where $$sin\theta\aprox \theta$$ then

$$F_{net}= T-Tsin \theta$$

I know that F= ma however you showed that

The restoring force is $Tsin\theta$ and if $theta$ is small and in radians then $sin\theta\approx \theta$

So that in the horizontal direction.

$$ma= -T\theta$$

and the distance it is displaced is given by the length of the arc (x) multiplied by the length,L, i.e. $\theta=xL$

$$ma= -T\theta$$

Well to show that something undergoes SHM, you need to show that $a=-\omega^2x$ where x=displacement,a=acceleration.

They should really tell us this but they do not and always ask to prove but never tell us what we are looking for...

Oh well I know that $$\omega = \sqrt{g} {L}$$

so... $$\theta= x L$$

then plugging in..

$$ma= -T\theta = ma$$

$$ma= -TxL$$

but do I just divide over the m? I'm not sure what that would accomplish for finding
$$a= \omega^2x$$

I know that $$\omega= \sqrt{g} {L}$$ and I'm not sure what relation I can find from this unfortunately.
Is there another one I need to think of to make it equal to having simple harmonic motion?

Thank you