Question

Show that under certain conditions, a simple pendulum performs linear S.H.M.

Answer 1

Let a mass m be suspended with a string of length l and be displaced by a force F through a small angle θ.

From figure, we see that the restoring force is mgsinθ.
Since the displacing force F is acting opposite to the restoring force
F = ─mgsin(θ)
Now since θ is small
sinθ ≈ θ
Let the horizontal displacement be x
θ = x/l
Thus,
F = ─mgθ
or, F = -mgx/l
or, F = -(mg/l)x
or, F = -kx  [since mg/l is a constant]
Restoring force is thus directly proportional to the displacement, hence the motion is SHM for small angles θ.

 or, F = -(mg/l)x
or, ma = -(mg/l)x
or, a = -(g/l)x

$a=-\omega^2x$ from condition of SHM, use it above,

$-\omega^2x=-(g/l)x$
$\omega^2=(g/l)$
$\omega=\sqrt{\frac{l}{g}}$
$T=2\pi\sqrt{\frac{l}{g}}$,
Here is clear that when displacement of pendulum is very small, then it follows simple harmonic motion. and then, time period of simple harmonic motion will be 2π$\sqrt{\frac{l}{g}}$

Answer 2

Answer:

}=2\pi\sqrt{\frac{L}{g}}

ANSWER