Proof dealing w/ Simple Harmonic Oscillations

[nix]

Prove that the period of a simple pendulum doing small oscillations is equal to:

2(py)x(square root of: l/g)

where py is 3.14..(obviously..lol)... l is length of the string of the pendulum and g is gravity

Also... the pendulum is basically just a ball on a string moving from side to side. and the equations we have been given to solve it is the fundamental eq. of an oscillation:

k = mw^2 = m x(2py^2/T) = m (2(py)f^2)

where f is the frequency, m is the mass, w is omega (the angle), and k is the spring constant and T is period

tricky..eh?

thanks for any help or suggestions..

 

[nix]

oops i just read the thing i was supposed to read before posting...

anyways this is what I've done and i don't think it makes sense maybe i should clarify the question with my teacher later..

m(2(py)^2/T) = m(2(py)f^2)

T = m(2(py)^2)/m(2(py)f^2)
T = py/f^2

f = 1/T

T = (py)T^2..maybe I am using the wrong equation...

And another thing about the question is that the angle is less than or equal to 5 degrees...the angle btw the original position of the pendulum and the position of the pendulum when its movind to the side.

 

[himanshu121]

T = py/f^2

f = 1/T

First equation has got wrong dimensions whereas second equation is correct.

Equations which might help u are
$$\omega^2 = \frac{k}{m}$$
$$\omega=2\pi f$$
$$T=\frac{2\pi}{\omega}$$
$$kl=mg$$

All equations are valid for small angle $$\theta\leq 5^0$$

 

[nix]

thanks so much for the help with the equations...i got it! yay!

but can you explain to me how kl = mg...

thanks again

 

[himanshu121]

That requires u to know the Force equation
i.e. $$F=-mgsin\theta$$
for small theta $$sin\theta = \theta$$
therefore the equation is $$F=-m\theta = -\frac{mgx}{l}$$
displacement along the arc=$$\theta l$$

also if u draw a free body diagram fo the forces . u have
$$k(l+\delta x)=mgcos\theta$$ for small displacement u can assume $$\delta x=0$$ & $$cos\theta=1$$
so u get kl=mg