In the dimensional equation for the period of oscillation of a simple pendulum, T = k g^a … — JAMB UTME

In the dimensional equation for the period of oscillation of a simple pendulum, $T = k g^a l^b$ , where $T$ is the period, $g$ is the acceleration due to gravity, $l$ is the length of the pendulum, and $k$ is a dimensionless constant, what is the numerical value of the exponent $a$ ?

Cevap: -0.5

Cevap

The numerical value of the exponent

a

is -0.5.

Applying dimensional analysis to

T = k g^a l^b

, the dimension of the left-hand side is

\text{T}^1

. The right-hand side has dimensions

(\text{L T}^{-2})^a (\text{L})^b = \text{L}^{a+b} \text{T}^{-2a}

. Equating the exponents of time

\text{T}

yields

1 = -2a

, which gives

a = -0.5

Adım Adım Çözüm

Express the dimensions of all physical quantities involved in fundamental base dimensions (M, L, T).

The dimension of period

T

[\text{T}]

, length

l

[\text{L}]

, and gravitational acceleration

g

[\text{L T}^{-2}]

Dimensional analysis requires substituting each quantity with its fundamental dimensions.

Formulate the dimensional homogeneity equation.

\text{M}^0 \text{L}^0 \text{T}^1 = (\text{L T}^{-2})^a (\text{L})^b = \text{L}^{a+b} \text{T}^{-2a}

The principle of dimensional homogeneity states that the exponents of base dimensions on both sides of a physically correct equation must be equal.

Equate exponents of time

\text{T}

and solve for

a

1 = -2a \implies a = -\frac{1}{2} = -0.5

Comparing the powers of

\text{T}

gives a linear equation in

a

Anahtar Kavram

Principle of Dimensional Homogeneity

Tahmini Süre:1m 15s

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