The period of a pendulum T depends on the mass of bob M length of the

1.	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» Answer: Given; $\longrightarrow \: \sf T\propto [M]^a [L]^b [g]^c\\$ Adding k as dimensionaless CONSTANT we have : $\longrightarrow \: \sf T = k [M]^a [L]^b [g]^c$ $\footnotesize\dag \: \bf{Dimensions \: of \: Time \: Period \: (T) = [M^0 L^0T ^{1} ]} \\$ $\footnotesize\dag\:\bf{Dimensions \: of \: mass \: (M) = [M^1 L^0 T^0]} \\$ $\footnotesize\dag\: \bf Dimensions \: of \: length \: (L) = [M^0L^1T^0]$ $\footnotesize\dag\: \bf Dimensions \: of \: Acceleration \: due \: gravity \: (g) = [M^0L^1T^{-2}] \\$ $\longrightarrow \: \sf [M^0L^0T^1] = [M^1 L^0 T^0]^a [M^0L^1T^0]^b [M^0L^1 T^{-2}]^c \\$ $\longrightarrow \: \sf [M^0L^0T^1] = [M]^a[L]^{b+ c} [ T]^{ - 2c }\\$ 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 : $\longrightarrow \: \gray{\underline{ \boxed{\orange{\rm T = 2\pi \sqrt{\frac{L}{g}}}}}}\\$

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»

Answer:

Given;

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

Adding k as dimensionaless CONSTANT we have :

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

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

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

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