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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 in air. Neglecting frictional force of water and given that the density of the bob is 3 kg/m3. What relationship between t and t0 is true?

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 in air. Neglecting frictional force of water and
given that the density of the bob is 3 kg/m3. What relationship between t and t0 is
true?

Grade:12

2 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
3 years ago
576-441_2.PNG
Arun
25750 Points
3 years ago

As we learnt in

Time period of oscillation of simple pendulum -

T=2\pi \sqrt{\frac{l}{g}}

- wherein

l = length of pendulum 

g = acceleration due to gravity.

 

 

 

t_{0}=2\pi \sqrt{l/g}.........(i)

Due to upthrust of water on the top, its apparent weight decreases

upthrust = weight of liquid displaced

\therefore \; \; \; Effective\; weight=mg-(V\sigma g)=V\rho g-V\sigma g

V\rho g`=Vg(\rho -\sigma ),where\; \sigma \; is \; density \; of\; water\;

or\; \; g`=g\left ( \frac{\rho -\sigma }{\rho } \right )

\therefore \; \; t=2\pi \sqrt{l/g`}=2\pi \sqrt{\frac{l\rho }{g(\rho -\sigma )}}..........(ii)

\therefore \; \; \; \frac{t}{t_{0}}=\sqrt{\frac{l\rho }{g(\rho -\sigma )}\times \frac{g}{l}}=\sqrt{\frac{\rho }{\rho -\sigma }}=\sqrt{\frac{4\times 1000/3}{(\frac{4000}{3}-1000)}}=2

or\; \; \; t=t_{0}\times 2=2t_{0}

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