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 t_0 in air. Negl - Physics MCQ AIEEE 2004 2026 | ExamTest.live

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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 t_{0} in air. Neglecting frictional force of water and given that the density of the bob is \frac{4}{3} \times 1000 kg/m^{3} . What relationship between t and t_{0} is true?

Solution & Step-by-step Explanation

Time period T \propto \frac{1}{g _{e ff}} .In air, g_{e ff} = g .In water, g_{e ff} = g - upthrust/mass = g (1 - \frac{ρ _{w a t er}}{ρ _{b o b}}) .Given ρ_{b o b} = \frac{4000}{3} kg/m^{3} and ρ_{w a t er} = 1000 kg/m^{3} . g_{e ff} = g (1 - \frac{1000}{4000/3}) = g (1 - \frac{3}{4}) = \frac{g}{4} Therefore, \frac{t}{t _{0}} = \frac{g}{g /4} = 4 = 2 . t = 2 t_{0} .

Try it yourself before checking the explanation above.

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 t_{0} in air. Neglecting frictional force of water and given that the density of the bob is \frac{4}{3} \times 1000 kg/m^{3} . What relationship between t and t_{0} is true?

t = t_{0}

t = t_{0} /2

t = 2 t_{0}

t = 4 t_{0}

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