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 surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms–2)

Answer 1

g E = 9.8 m/s
g M = 1.7 m/s
t E = 3.5 s
t M = ?

g = v - u /t
9.8 = v - 0 /3.5
9.8 × 3.5 = v
34.3=v

now;
gM /v= 1/t
34.3/ 1.7 = t
= 20.17 seconds.

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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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there fore 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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