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
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∵ 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 %
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 %
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12
Answer: how come the time period is 90 seconds
Time period= no. Of oscillations completed per second i.e, T = 50/40 s
Explanation:
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