Question

The period of a pendulum T depends on the mass of bob M length of the trends L and the acceleration due to gravity G use period of dimensions to find an expression for T

Answer 1

Answer:

Given;

$\longrightarrow \: \sf T\propto [M]^a [L]^b [g]^c$

Adding k as dimensionaless constant we have :

$\longrightarrow \: \sf T = k [M]^a [L]^b [g]^c$

$\footnotesize\dag \: \bf{Dimensions \: of \: Time \: Period \: (T) = [M^0 L^0T ^{1} ]}$

$\footnotesize\dag\:\bf{Dimensions \: of \: mass \: (M) = [M^1 L^0 T^0]}$

$\footnotesize\dag\: \bf Dimensions \: of \: length \: (L) = [M^0L^1T^0]</p><p></p><p>$$\footnotesize\dag\: \bf Dimensions \: of \: Acceleration \: due \: gravity \: (g) = [M^0L^1T^{-2}]$

$\longrightarrow \: \sf [M^0L^0T^1] = [M^1 L^0 T^0]^a [M^0L^1T^0]^b [M^0L^1 T^{-2}]^c$

$\longrightarrow \: \sf [M^0L^0T^1] = [M]^a[L]^{b+ c} [ T]^{ - 2c }$

On comparing the powers of LHS and RHS we get:

$\dashrightarrow\sf a = 0$

$\dashrightarrow\sf b + c= 0$

$\dashrightarrow\sf b= - c \qquad \: ...(i)$

$\dashrightarrow\sf -2c= 1$$\dashrightarrow\sf c= \dfrac{1}{- 2} \qquad \: ...(ii)$

Substituting the value of c from eqⁿ (ii) to eqⁿ (i) we get :

$\dashrightarrow\sf b = - \bigg(\dfrac{1}{-2} \bigg)$

$\dashrightarrow\sf b = \dfrac{1}{2}$$\longrightarrow \: \sf T = k [M]^0 [L]^{ \frac{1}{2} } [g]^{ \frac{1}{ - 2} }$

$\longrightarrow \: \sf T = k [L]^{ \frac{1}{2} } [g]^{ \frac{1}{ - 2}}$

$\longrightarrow \: \sf T = k \dfrac{[L]^{ \frac{1}{2} } }{[g]^{ \frac{1}{2}}}$

$\longrightarrow \: \sf T = k \dfrac{\sqrt{[L] }}{\sqrt{[g]}}$

$\longrightarrow \: \sf T = k \sqrt{ \frac{L}{g} }$

Experimentally we know that the value of k is 2π, Hence :

$\longrightarrow \: \gray{\underline{ \boxed{\orange{\rm T = 2\pi \sqrt{\frac{L}{g}}}}}}$