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Oscillations — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsOscillations4 Marks
Question
Define a seconds pendulum. Find an expression for its length at a given place. Show that the length of a seconds pendulum has a fixed value at a given place.
✓

Answer


(1) Seconds pendulum: A simple pendulum of period two seconds is called a seconds pendulum.
(2) The period of a simple pendulum is
$
T =2 \pi \sqrt{\frac{L}{g}}
$
For a seconds pendulum, $T =2 s$.
$
\therefore 2=2 \pi \sqrt{\frac{L}{g}} \therefore L =\frac{g}{\pi^2}
$
This expression gives the length of the seconds pendulum at a place where acceleration due to gravity is $g$.
(3) At a given place, the value of $g$ is constant.
$\therefore L = g / \pi^2=$ a fixed value, at a given place.
[Note: Because the effective gravitational acceleration varies from place to place, the length of a seconds pendulum should be changed in direct proportion. Since the effective gravitational acceleration increases from the equator to the poles, so should the length of a seconds pendulum be increased.]

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