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Karnataka Board PUCPUC Science Class 11

A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period.

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Question

A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period. `T = 2πsqrt(m/(Apg))` where m is mass of the body and ρ is density of the liquid.

Long Answer
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Solution

Consider the diagram,


Let the log be pressed and let the vertical displacement at the equilibrium position be x0.

At equilibrium, mg = buoyant force = (ρAx0)g  ......[∵ m = Vρ = (Ax0)ρ]

When it is displaced by a further displacement x, the buoyant force is A(x0 + x)ρg

∴ Net restoring force = Buoyant force – Weight

= A(x0 + x)ρg – mg

= (Aρg)x   .....[∵ mg = ρAx0g]

As displacement x is downward and restoring force is upward, We can write 

`F_"restoring" = - (Aρg)x`

= `- kx`

Where k = constant = Aρg

So, the motion is S.H.M  .....(∵ F ∝ – x)

Now, Acceleration a = `F_"restoring"/m = - k/m x`

Comparing with a = `- ω^2x`

⇒ `ω^2 = k/m`

⇒ `ω = sqrt(k/m)`

⇒ `(2pi)/T = sqrt(k/m)`

⇒ `T = 2pi sqrt(m/k)`

⇒ `T = 2pi sqrt(m/(Aρg))`

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Chapter 14: Oscillations - Exercises [Page 104]

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NCERT Exemplar Physics Exemplar [English] Class 11
Chapter 14 Oscillations
Exercises | Q 14.37 | Page 104

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