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  • Need explanation for: he 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 air. Neglecting frictional force of water and given that the density of the bob is (4/3

he 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 air. Neglecting frictional force of water and given that the density of the bob is (4/3) x 1000 kg/m3. What relationship between t and t0  is true ?

  • Option 1)

    t = t0

  • Option 2)

    t = t0/2

  • Option 3)

    t = 2t0

  • Option 4)

    t = 4t0

 

Answers (1)

best_answer

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}

Correct option is 3.


Option 1)

t = t0

This is an incorrect option.

Option 2)

t = t0/2

This is an incorrect option.

Option 3)

t = 2t0

This is the correct option.

Option 4)

t = 4t0

This is an incorrect option.

Posted by

Aadil

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