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Oscillations

Question
CBSEENPH11020484

The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t0 in the air. Neglecting frictional force of water and given that the density of the bob is (4/3) x 1000 ms-1 . What relationship between t and t0 is true?

  • t = t0

  • t = t0/2

  • t = 2t0

  • t = 4t0

Solution

C.

t = 2t0

The time period of simple pendulum in air
straight T space equals space straight t subscript 0 space equals space 2 straight pi space square root of 1 over straight g end root space...... space left parenthesis straight i right parenthesis
being the length of simple pendulum, In water, effective weight of bob
w' = weight of bob in air - upthrust
⇒ ρVgeff = mg-m'g = ρVg-ρ'Vg = (ρ-ρ') where ρ = density of bob, ρ′ = density of water
straight g subscript eff space equals space open parentheses fraction numerator straight rho minus straight rho apostrophe over denominator straight rho end fraction close parentheses space straight g
space equals space open parentheses 1 minus fraction numerator straight rho apostrophe over denominator straight rho end fraction close parentheses straight g
therefore space
space straight t equals space 2 straight pi space square root of open square brackets fraction numerator 1 over denominator open parentheses 1 minus begin display style fraction numerator straight rho apostrophe over denominator straight p end fraction end style close parentheses end fraction close square brackets end root space........ space left parenthesis ii right parenthesis
Thus comma space straight t over straight t subscript straight o space equals space square root of open square brackets fraction numerator 1 over denominator 1 minus begin display style fraction numerator straight rho apostrophe over denominator straight rho end fraction end style end fraction close square brackets end root
space equals space square root of open parentheses fraction numerator 1 over denominator 1 minus begin display style fraction numerator 1000 over denominator left parenthesis 4 divided by 3 right parenthesis space straight x space 1000 end fraction end style end fraction close parentheses end root space equals space square root of open parentheses fraction numerator 4 over denominator 4 minus 3 end fraction close parentheses end root

space equals space 2 rightwards double arrow space straight t space space equals 2 straight t subscript 0