Class 11 Physics Oscillations | Simple Pendulum |

__Simple Pendulum__

A simple pendulum is defined as an object that has a small mass (pendulum bob), which is suspended from a wire or string having negligible mass.

- Whenthe pendulum bob is displaced it oscillates on a plane about the vertical line through the support.
- Simple pendulum can be set into oscillatory motion by pulling it to one side of equilibrium position and then releasing it.

In the above image one end of a bob of mass m is attached to a string of length L and another to a rigid support executing simple harmonic motion.

__Problem__**:-**What is the length of a simple pendulum, which ticks seconds?

**Answer:**-

T =√ L /g

By using above formula

L = gT^{2}/4^{2}

The time period of a simple pendulum, whichticks seconds, is 2 s. Therefore, for g=9.8 m s^{–2}

and T = 2 s,

L = (9.8(m s^{-2}) 4(s^{2}))/4^{2}= 1 m

** Problem:-**A cylindrical piece of cork of density of base area A and height h floats in a liquid ofdensity ρ

T = 2 π√hρ/ρ_{l}g

where ρ is the density of cork. (Ignore damping due to viscosity of the liquid)

__Answer:-__

Base area of the cork = A

Height of the cork = h

Density of the liquid = ρ1

Density of the cork = ρ

In equilibrium:

Weight of the cork = Weight of the liquid displaced by the floating cork

Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.

Up-thrust = Restoring force, F = Weight of the extra water displaced

F = –(Volume × Density × g)

Volume = Area × Distance through which the cork is depressed

Volume = Ax

∴ F = – A x ρ1 g .....(i)

According to the force law:

F = kx

k = F/x

where, k is constant

k = F/x = -Aρ1 g.... (ii)

The time period of the oscillations of the cork:

T = 2π√m/k .... (iii)

where,

m = Mass of the cork

= Volume of the cork × Density

= Base area of the cork × Height of the cork × Density of the cork

= Ahρ

Hence, the expression for the time period becomes:

T=2π√Ahρ/Ahρ_{1}g

T = 2π√hρ/ρ_{1}g

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