Answer:
Length of longer pendulum = 99.3 cm
Length of shorter pendulum = 82.2 cm
Step-by-step explanation:
Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.
From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²
substituting T = 2s and g = 9.8 m/s² we have
l = T²g/4π²
= (2 s)² × 9.8 m/s² ÷ 4π²
= 39.2 m ÷ 4π²
= 0.993 m
= 99.3 cm
Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.
So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.
t/T₂ = t/T₁ + x
1/T₂ = 1/T₁ + x (1)
Also, the T₂ = t/n₂ = t/(n₁ + x) (2)
From (1) T₂ = T₁/(T₁ + x) (3)
equating (2) and (3) we have
t/(n₁ + x) = T₁/(T₁ + x)
substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have
20 s/(10 + x) = 2/(2 + x)
10/(10 + x) = 1/(2 + x)
(10 + x)/10 = (2 + x)
(10 + x) = 10(2 + x)
10 + x = 20 + 10x
collecting like terms
10x - x = 20 - 10
9x = 10
x = 10/9
x = 1.11
x ≅ 1 oscillation
substituting x into (2)
T₂ = t/n₂ = t/(n₁ + x)
= 20/(10 + 1)
= 20/11
= 1.82 s
Since length l = T²g/4π²
substituting T = 1.82 s and g = 9.8 m/s² we have
l = T²g/4π²
= (1.82 s)² × 9.8 m/s² ÷ 4π²
= 32.46 m ÷ 4π²
= 0.822 m
= 82.2 cm