Physics, asked by sweety105, 1 year ago

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)

Answers

Answered by Hriday0102
8
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.

PLZ MARK MY ANSWER AS BRAINLIEST AS SOON AS POSSIBLE IF U THINK IT IS CORRECT!!!!!!!!!!!!

Hriday0102: Is my answer correct??
sweety105: no ans is 8.4 s
Hriday0102: ok sorry for the wrong answer
Hriday0102: Can u check where I m wrong
sweety105: u check
Hriday0102: ok
Hriday0102: Actually I have tried this type of question the first time
Hriday0102: That is of which class
sweety105: it's of 11th
Answered by TheRuhanikaDhawan
7
on earth
---------------------------------
T = 2π √ l/g
-------------
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
-------------------------------
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
----------------------------------------------------------------------------------------
there fore the time period of a simple pendulum on the surface of moon will be 8.35 s  or approximately = 8.4 s
---------------
Similar questions