Question

#Moderatorschallenge

This is working of simple Pendulum. pls explain me how T is proportional to √l and ✓1/g.
Prove that!

There is graph on left hand
side.

NOTE
•t-Time period
•l-length
•g-gravity

what I don't understand is that why they have taken sq root in both cases??
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Answer 1

Answer:

SIMPLE PENDULUM –

Let us suppose "t" as the time period taken by the pendulum for one ossicilation. "l" is length of the string. And "g" as the gravitational force.

We know, "t" is directly proportional to √l and indirectly proportional to √g.

Hence, t ∝ √l and t ∝√(1/g).

Hence, t ∝ √(l/g)

Now, t = k √(l/g)

Hence, t = 2π √(l/g)

2π is the angle in radians subtended by a whole circle of 360° which is travelled by the bob in two terms of 180° and 180° (a cycle).

By dimensional analysis –

The dependence of time period for oscillation on the factors l,g and m i.e. length, gravity and mass can be represented as :

➸ T = k {l}^x {g}^y {m}^z

Here k is dimensionless quantity and x, y and z are exponents.

Considering dimensions on both sides :

➸ [L⁰M⁰T¹] = [L¹]^x [L¹ T^{-2}]^y [M¹]^z

➸ [L⁰M⁰T¹] = L^x L^y T^{-2y} M^z

➸ [L⁰M⁰T¹] = L^{x+y} T^{-2y} M^z

Equating dimensions on both sides,

➸ x+y = 1

➸ -2y = 1

➸ z = 0

We get, y = -½, x = ½ and z = 0.

Hence, T = k l^{½} g^{-½}

➸ T = k √(l/g)

➸ T = 2π √(l/g)