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 .
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Answer 1

★ Concept :-

Here the concept of Dimensional Analysis of Equation has been used. We know that by the use of dimensions of seven base physicsl quantities, we can deduce a formula or equation by using the relations. We will take equation using a constant derived through the proportional symbol.

Let's do it !!

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★ Solution :-

Given,

Time Period of oscillation of pendulum depends on,

» mass of the bob (m)

» length of the pendulum (l)

» acceleration due to gravity at that place (g)

• Let the Time Period of the Simple Pendulum be T

We know that since the Time Period is dependent to these qualities. So we can use proportionality symbol here.

Thus we get our equation as,

$\\\;\bf{\mapsto\;\;\green{T\;\;\propto\;\;m^{a}\:l^{b}\:g^{c}}}$

Here a, b and c are the powers to which m, l and g are raised which makes T depend on them.

The above equation can be written as,

$\\\;\bf{\mapsto\;\;\red{T\;\;=\;\;K\:m^{a}\:l^{b}\:g^{c}}}$

where K is the proportionality constant.

Also this K is dimensionless because its value is to get equality in the equation.

Let's keep this main equation aside for sometimes.

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~ For Dimensions of different quantities ::

→ For dimensions of T ::

We know that T is the Time Period so its dimension will be the standard dimension of time. So according to dimensional analysis we get,

✒ [ T ] = T

$\;\tt{\blue{\rightarrow\;\;[T]\;=\;M^{0}\:L^{0}\:T^{1}}}$

Here T also shows the standard dimension of Time.

→ For dimension of m ::

We know that m is the mass of the bob so its dimension will be the standard dimension of mass. So, according to dimensional analysis we get,

✒ [m] = M

$\;\tt{\blue{\rightarrow\;\;[m]\;=\;M^{1}\:L^{0}\:T^{0}}}$

• For dimensions of l ::

We know that l is the length of the bob so its dimensions will be the standard dimension of length. So, according to the dimensional analysis we get,

✒ [ l ] = L

$\;\tt{\blue{\rightarrow\;\;[l]\;=\;M^{0}\:L^{1}\:T^{0}}}$

• For dimensions of g ::

We know that g is the acceleration due to gravity so its unit will be m/sec² . Thus using the method of dimensional analysis we get,

✒ [g] = L / T²

$\;\tt{\blue{\rightarrow\;\;[g]\;=\;M^{0}\:L^{1}\:T^{-2}}}$

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~ For the Derivation of Expression ::

Using the dimensions of different quantities in the main equation, we get

$\\\;\bf{\mapsto\;\;T\;\;=\;\;K\:m^{a}\:l^{b}\:g^{c}}$

Now by applying the method of dimensional analysis, we get

$\\\;\sf{\Longrightarrow\;\;[T]\;\;=\;\;\bf{K\:[m]^{a}\:[l]^{b}\:[g]^{c}}}$

Since, a, b and c are exponents so they will be dimensionless constants like K.

By applying the dimensions of different quantities, we get

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

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

By using laws of exponents, we get

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

Now removing the dimensional analysis from both sides, we get

$\\\;\sf{\Longrightarrow\;\;M^{0}\:L^{0}\:T^{1}\;\;=\;\;\bf{K\:M^{a}\:L^{b}\:L^{c}\:T^{-2c}}}$

Let's rearrange the equation and use the laws of exponents. Using that, we get

$\\\;\sf{\Longrightarrow\;\;M^{0}\:L^{0}\:T^{1}\;\;=\;\;\bf{K\:M^{a}\:L^{b\:+\:c}\:T^{-2c}}}$

We see that both sides we have same quantities but with different exponents. So comparing only exponents of LHS and RHS, we get

→ Values of equations,

✒ a = 0 , b + c = 0 , -2c = 1

On solving this, we get

✒ a = 0, b = -c , c = - ½

✒ a = 0 , b = - (- ½) , c = - ½

✒ a = 0, b = ½ , c = - ½

Now applying these values of exponents in the main equation which we got earlier, we get

$\\\;\bf{\Longrightarrow\;\;T\;\;=\;\;K\:m^{0}\:l^{1/2}\:g^{-\:1/2}}$

Since exponent as 0 has value as 1 which on multiplication gives same value. So let's remove it now.

$\\\;\bf{\Longrightarrow\;\;T\;\;=\;\;K\:l^{1/2}\:g^{-\:1/2}}$

$\\\;\bf{\Longrightarrow\;\;T\;\;=\;\;K\:l^{1/2}\:\dfrac{1}{g^{1/2}}}$

$\\\;\bf{\Longrightarrow\;\;T\;\;=\;\;K\:\dfrac{l^{1/2}}{g^{1/2}}}$

Again using the laws of exponents, we get

$\\\;\bf{\Longrightarrow\;\;T\;\;=\;\;K\:\sqrt{\dfrac{l}{g}}}$

From experimental calculations, we know the value of K as 2π . So,

$\\\;\bf{\Longrightarrow\;\;T\;\;=\;\;\orange{2\pi\:\sqrt{\dfrac{l}{g}}}}$

This is the required expression.

$\\\;\underline{\boxed{\tt{Required\:\;derivation\;\:of\;\:expression\;=\;\bf{\purple{2\pi\:\sqrt{\dfrac{l}{g}}}}}}}$

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★ More to know :-

• Dimension of mass = [M]

• Dimension of Time = [T]

• Dimension of amount of substance = [mol]

• Dimension of Length = [L]

• Dimension of electric current = [A]

• Dimension of luminous intensity = [cd]

• Dimension of temperature = [K]

Answer 2

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HOPE IT HELPS