The period for small oscillations of a system

1. May 30, 2014

zeralda21

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

See picture :

2. Relevant equations

$\sum M_{O}=I_{O}\ddot{\theta }$

3. The attempt at a solution

Consider the free-body diagram associated with an arbitrary positive angular displacement $\theta$; The moment about point $O$ is given by

$\sum M_{O}=-k\left ( b\sin\theta \right )\left ( b\cos\theta \right )-k\left ( 2b\sin\theta \right )\left ( 2b\cos\theta \right )-\underbrace{\left ( 3mgb\sin\theta \right )\left ( 3b\cos\theta \right )}_{\mathit{Why \ shouldn't \ this \ be \ included?}}$

Further, by the parallel axis theorem, $I_{0}=\overline{I}+md^2=0+m(3b)^2=9mb^2$ and for small oscillations $\sin\theta\simeq \theta \ \ \wedge \cos\theta\simeq 1$ and $\tau =\frac{2\pi}{\omega _{n}}=2\pi\sqrt{\frac{m}{k}}$. BUT why does not the mass of the sphere contribute to the moment?

2. May 30, 2014

ehild

The springs are stretched initially to keep the rod and mass horizontal, so the torque of gravity is balanced by the torque of the tensions in the springs. You only need to take into account the additional torques if you displace the mass from the equilibrium. The torque of gravity does not change if the displacement is small as it is proportional to the cosine of the angle.

ehild

3. May 31, 2014

Tanya Sharma

Hello ehild

Why should both the springs be stretched in the equilibrium position ? Either the upper one should be stretched or the lower one should be compressed or may be both the conditions hold simultaneously .

Am I right ?

Let the lower spring be compressed by x1 and upper stretched by x2 in equilibrium .Then,

For translational equilibrium , kx1+kx2 = mg

For rotational equilibrium kx1b+kx2(2b) = mg(3b)

Solving the above ,we get x2 = 2mg/k and x1 = -mg/k .

4. May 31, 2014

ehild

By "stretched" I meant that the springs are under tension (well, not necessarily both). You do not know the relaxed lengths. And it can happen that both are really stretched.

That is not true. You ignored the force of the pivot.

That is correct.
As the first equation is not valid you can not determine x1 and x2.

5. May 31, 2014

Tanya Sharma

As always you are right .

Writing torque equation,

$$M(3b)^2\ddotθ = -[k(x_1+bsinθ)(bcosθ)+k(x_2+2bsinθ)(2bcosθ) - mg(3b)]$$

Now $kx_1(bcosθ)+kx_2(2bcosθ) ≈ mg(3b)$

$$9Mb^2\ddotθ = -[k(bsinθ)(bcosθ)+k(2bsinθ)(2bcosθ)]$$

Applying approximations ,

$$9Mb^2\ddotθ = -[k(bsinθ)(bcosθ)+k(2bsinθ)(2bcosθ)]$$

$$9Mb^2\ddotθ = -5kb^2θ$$

$$\ddotθ = -\frac{5k}{9M}θ$$

$$\omega = \sqrt{\frac{5k}{9M}}$$

$$T = 2\pi \sqrt{\frac{9M}{5k}}$$

Is this correct ?

6. May 31, 2014

ehild

It looks good

ehild

7. May 31, 2014

Thanks .