Question

time period (T) of oscillation of simple pendulum depends on length T . mass m acceleration due to gravity g deduce the relation. among them using the method of dimensions​

Answer 1

Topic :- Units and Dimensions

$\maltese\:\underline{\sf AnsWer :}\:\maltese$

Let the Time Period (T) depends on mass (M) , length (L) and and Acceleration due to gravity (g)

$\longrightarrow \: \sf T\propto [M]^x [L]^y [g]^z$

Adding k as proportionality constant we have :

$\longrightarrow \: \sf T = k [M]^x [L]^y [g]^z$

Now,

$\dag \: \textsf{Dimensions of Time Period (T) = [T]}$

$\dag\:\textsf{Dimensions of mass (M) = [M]}$

$\dag\:\textsf{Dimensions of length (L) = [L]}$

$\dag\:\textsf{Dimensions of Acceleration due gravity (g) = [LT ^\text{-2} ]}$

Hence,

$\longrightarrow \: \sf [M^0L^0T^1] = [M^1 L^0 T^0]^x [M^0L^1T^0]^y [M^0L^1 T^{-2}]^z$

$\longrightarrow \: \sf [M^0L^0T^1] = [M]^x [L]^{y + z} [ T]^{ - 2z }$

On comparing the powers of LHS and RHS we get:

$\dashrightarrow\sf x = 0$

Now,

$\dashrightarrow\sf y + z = 0$

$\dashrightarrow\sf y = - z \qquad \: ...(i)$

Also,

$\dashrightarrow\sf -2z= 1$

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

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

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

$\dashrightarrow\sf y = \dfrac{1}{2}$

Hence,

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

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

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

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

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