Question

The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy. The time for 50 oscillations was measured to be 40 s within 1 s resolution. Calculate the percentage accuracy in the determination of acceleration due to gravity ‘g’ from the above measurement

Answer 1

Given conditions ⇒ 
∆l = 1mm
 = 0.1 cm 
l = 20cm 

∆t = 1 s
and t = 90 s 

Solution is shown below, 

We know that ⇒
 ∆T/T = ∆t/t 

As per as the Formula,
 T = 2π√(l/g) 
On squaring on both the sides, 
T² = 4π²l/g
∴ g = 4π²l/T² 
⇒ ∆g/g = ∆l/l + 2∆T/T
⇒ ∆g/g = ∆l/l + 2∆t/t [ from ∆T/T = ∆t/t   ]

Thus, 
% error in g = 100( ∆l/l + 2∆t/t) 
∴ % error in g = 100(0.1/20 + 2 × 1/90)
= 100( 0.005 + 0.022)
= 100( 0.027)
= 2.7 % 

Hence, the % error in the acceleration due to the gravity is 2.7 %

Answer 2

∵ time period is time taken to complete a revolution .
e.g., T = t/n , here n is number of revolution and t is time.
so, ∆T = ∆t/n, because n is Constant term.
Hence, ∆T/T = ∆t/n × n/t = ∆t/t -----(1)

Now, T = 2π√(l/g)
Taking square both sides,
T² = 4π²l/g ⇒ g = 4π²l/T²
∴ ∆g/g = ∆l/l + 2∆T/T
⇒ ∆g/g = ∆l/l + 2∆t/t [ from equation (1) ]
∴ % error in g = 100( ∆l/l + 2∆t/t)
Here, ∆l = 1mm = 0.1 cm , l = 20cm
∆t = 1 s and t = 90 s
∴ % error in g = 100(0.1/20 + 2 × 1/90)
= 100( 0.005 + 0.022)
= 100( 0.027)
= 2.7 %

Hence, % error in g = 2.7 %