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.

where m is mass of the body and ρ is density of the liquid​

Answer 1

Hence proved that the wood execute S.H.M. with a time period T = 2π √ m /Apg

Explanation:

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

mg=buoyant force=(qAx0)g [∴m=vp=(Ax0)p]

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

Net restoring force=Buoyant force−Weight

Net restoring force = A(x0+x)pg−mg  

Net restoring force = (Apg)x [∴mg=pAx0g]

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

F(restoring) = − (Apg)x

Where k = constant = Apg

So, the motion is SHM (∴F∝−x)

Now, Acceleration a = F(restoring) m = −k / m x

Comparing with a = −ω^2.x

⇒ ω^2 = k/m ⇒ ω = √k / m

⇒ 2π / T = √k / m ⇒ T = 2π √m / k

T = 2π √ m /Apg

Hence proved that the wood execute S.H.M. with a time period T = 2π √ m /Apg .

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Answer 2

Answer:

• Time period is T = 2π √ m /Apg

Given:

1. Area of the log of wood = A
2. Height of the log of wood = h
3. Density of Liquid = ρ

Explanation:

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Now we know that, in a simple harmonic motion:

⟼ F is directly proportional to -x

⟼ F ∝ - x

⟼ F = - kx, where k is a constant

Also we know that,

⟼ F = mg

But since the force is upthrust force, the force will be opposite and equal in magnitude therefore we denote it as - mg.

⟼ F = - mg

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Now, we know that,

⟼ Density = Mass / Volume

⟼ Mass = Density × Volume =  ρ × V

⟼ Volume = Area × Displaced Length = Ax

Substituting,

⟼ Mass of the water displaced =  ρAx

Now,

⟼ Force (F) = - ρAx × g = - ρgAx

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Now we also know that,

⟼ F = - kx

⟼ k = - F/x

⟼ k = - ( - ρgAx ) / x

⟼ k =  ρgA

Now for an object exhibiting Simple Harmonic Motion (SHM), we know that,

⟼ T = 2π √ (m/k )

Substituting " k " value in the given equation. we get,

⟼ T = 2π √ ( m )/(ρgA)

∴ Time period is T = 2π √ ( m )/(ρgA)

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