First slide

Dimensions of physical quantities

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 oscillations of the simple pendulum depends on its length (l ), mass of the bob (m) and acceleration due to gravity (g). Using the method of dimensions, expression for its time period is

Moderate

A

$T \propto k \sqrt{\frac{g}{l}}$
B

$T \propto k \sqrt{\frac{2 g}{l}}$
C

$T \propto k \sqrt{\frac{l}{g}}$
D

$T \propto k \sqrt{\frac{l}{2 g}}$

Solution

The dependence of time period T on the quantities l, g and mass a product may be written as $T = k l^{x} g^{y} m^{z}$
where, k is dimensionless constant and x, y and z are the exponents. Taking dimensions on both sides, we have
$[L^{0} M^{0} T^{- 1}] = {[L^{1}]}^{x} {[T^{- 2}]}^{y} {[M^{1}]}^{z}$
$[M^{0} L^{0} T^{1}] = M^{z} L^{x + y} T^{- 2 y}$
On equating the dimensions on both sides, we have
x + y = 0,
-2y =1
$\Rightarrow y = - \frac{1}{2} a n d x = \frac{1}{2}$
and z = 0
So that $T = k l^{1 / 2} g^{- 1 / 2}$
or $T = k \sqrt{\frac{l}{g}}$
NOTE:
The value of constant k cannot be obtained by the method of dimensions. Here, it does not matter if some number multiplies the right side of this formula, because that does not affect its dimensions.
Actually, $k = 2 π s o t h a t T = 2 π \sqrt{\frac{l}{g}}$

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