Question

The acceleration due to gravity on the surface of moon is 1.7 ms-2. What is the time period of
a simple pendulum on the moon if it time period is 3.5 s on earth

Answer 1

On earth T=3.5 S
g=9.8 m/s²

T=2π√(l/g)
3.5=2π√(l/9.8)

On moon :
T1=?
g1=1.7 m/s²

T1=2π√(l/g)
=2π√(l/1.7)

⇒T1/3.5=√(9.8/1.7)

T1=240x3.5=8.4sec

Answer 2

On Earth
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T = 2π √ l/g
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we know that T = 3.5 s

3.5 = 2π √ l/g

taking sq on both sides

(3.5)² = (2π)² x l/g

          T²
  l = ----------- x g
         (2π)²

           (3.5)²
l    =  ---------     x 9.8
          (6.28)²

solving this we get length = 3.04 m
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on moon substituting the length in the time period formula we get

T = 2π √ l/g

T = 2π ( √ 3.04/1.7)

T = 2π (√1.7)

T = 6.28 x 1.33

T = 8.35 s
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Therefore the time period of a simple pendulum on the surface of moon will be 8.35 s  or approximately = 8.4 s
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