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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. |
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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 = 2π \(\sqrt{\frac Lg}\) For a seconds pendulum, T = 2s. ∴ 2 = 2π\(\sqrt{\frac Lg}\) ∴ L = \(\frac{g}{π^2}\) 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. ∴ L = g/π2 = 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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