Q.(a) Obtain an expression for the time period T of a simple pendulum. The time period depends on :
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Start your 14-day free trial to unlock the full solution →Assuming T = k m^a l^b g^c and matching dimensions on both sides gives a = 0, b = 1/2, c = -1/2, so T = 2 pi sqrt(l/g).
This question offers a choice between (a) deriving the time period of a simple pendulum by the dimensional method, and (b) the triangle law of vector addition; part (a) is answered here.
Let the time period T of a simple pendulum depend on the mass m of the bob, the length l of the pendulum, and the acceleration due to gravity g, as a power-law relation:
T = k m^a l^b g^c ... (1)
where k is a dimensionless constant (given as k = 2 pi).
Writing the dimensions of each quantity:
[T] = T^1
[m] = M^1
[l] = L^1
[g] = L^1 T^-2
Substituting into equation (1):
M^0 L^0 T^1 = M^a L^b (L T^-2)^c = M^a L^(b+c) T^(-2c)
Comparing powers of M, L, T on both sides:
Power of M: 0 = a gives a = 0
Power of T: 1 = -2c gives c = -1/2
Power of L: 0 = b + c gives b = -c = 1/2
So the relation becomes
T = k m^0 l^(1/2) g^(-1/2) = k sqrt(l/g)
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