Question

Consider a simple pendulum, having a bob attached to a string , that oscillates under the action of the force of gravity. Suppose,that the period of oscillation of pendulum depends on

i) mass of the bob(m)

(ii) length of the pendulum(l)

(iii ) acceleration due to gravity at the place.​(g)

• Derive the expression for its time period using method of dimensions .
$➝\blue{\underline{\boxed{\sf Don't \: attach \: image \: need \: typed \: answer:}}}$
​

Answer 1

Let the period $\sf{T}$ depend on mass of the bob $\sf{m,}$ length of the pendulum $\sf{l}$ and acceleration due to gravity $\sf{g}$ as the following.

$\sf{\longrightarrow T\propto m^x\,l^y\,g^z}$

or,

$\sf{\longrightarrow T=k\cdot m^x\,l^y\,g^z\quad\quad\dots(1)}$

where $\sf{k}$ is a dimensionless constant.

Taking dimension of each term in the equation,

$\sf{\longrightarrow [T]=[m]^x[l]^y[g]^z}$

$\sf{\longrightarrow T=M^x\,L^y\,\left(LT^{-2}\right)^z}$

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

Equating powers of corresponding terms in both sides of the equation, we get,

• $\sf{x=0}$

• $\sf{y+z=0}$

• $\sf{-2z=1}$

Solving them we get,

• $\sf{x=0}$

• $\sf{y=\dfrac{1}{2}}$

• $\sf{z=-\dfrac{1}{2}}$

Then (1) becomes,

$\sf{\longrightarrow T=k\cdot m^0\,l^{\frac{1}{2}}\,g^{-\frac{1}{2}}}$

$\sf{\longrightarrow\underline{\underline{T=k\sqrt{\dfrac{l}{g}}}}}$

Experimentally, $\sf{k=2\pi}$ here. Hence,

$\sf{\longrightarrow\underline{\underline{T=2\pi\sqrt{\dfrac{l}{g}}}}}$

Answer 2

Let the period \sf{T}T depend on mass of the bob \sf{m,}m, length of the pendulum \sf{l}l and acceleration due to gravity \sf{g}g as the following.

\sf{\longrightarrow T\propto m^x\,l^y\,g^z}⟶T∝m

x

l

y

g

z

or,

\sf{\longrightarrow T=k\cdot m^x\,l^y\,g^z\quad\quad\dots(1)}⟶T=k⋅m

x

l

y

g

z

…(1)

where \sf{k}k is a dimensionless constant.

Taking dimension of each term in the equation,

\sf{\longrightarrow [T]=[m]^x[l]^y[g]^z}⟶[T]=[m]

x

[l]

y

[g]

z

\sf{\longrightarrow T=M^x\,L^y\,\left(LT^{-2}\right)^z}⟶T=M

x

L

y

(LT

−2

)

z

\sf{\longrightarrow M^0\,L^0\,T^1=M^x\,L^{y+z}\,T^{-2z}}⟶M

0

L

0

T

1

=M

x

L

y+z

T

−2z

Equating powers of corresponding terms in both sides of the equation, we get,

\sf{x=0}x=0

\sf{y+z=0}y+z=0

\sf{-2z=1}−2z=1

Solving them we get,

\sf{x=0}x=0

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

2

1

\sf{z=-\dfrac{1}{2}}z=−

2

1

Then (1) becomes,

\sf{\longrightarrow T=k\cdot m^0\,l^{\frac{1}{2}}\,g^{-\frac{1}{2}}}⟶T=k⋅m

0

l

2

1

g

−

2

1

\sf{\longrightarrow\underline{\underline{T=k\sqrt{\dfrac{l}{g}}}}}⟶

T=k

g

l

Experimentally, \sf{k=2\pi}k=2π here. Hence,

\sf{\longrightarrow\underline{\underline{T=2\pi\sqrt{\dfrac{l}{g}}}}}⟶

T=2π

g

l