SATHEE: Simple Harmonic Motion 3 Question 11

Simple Harmonic Motion 3 Question 11

11. A thin rod of length $L$ and uniform cross-section is pivoted at its lowest point $P$ inside a stationary homogeneous and non-viscous liquid. The rod is free to rotate in a vertical plane about a horizontal axis passing through $P$ .

The density $d_{1}$ of the material of the rod is smaller than the density $d_{2}$ of the liquid. The rod is displaced by small angle $θ$ from its equilibrium position and then released. Show that the motion of the rod is simple harmonic and determine its angular frequency in terms of the given parameters.

$(1996, 5$ M)

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Answer:

Correct Answer: 11. $ω = \sqrt{\frac{3 g (d_{2} - d_{1})}{2 d_{1} L}}$

Solution:

Let $S$ be the area of cross-section of the rod. In the displaced position, as shown in figure, weight $(w)$ and upthrust $(F_{B})$ both pass through its centre of gravity $G$ . Here, $w = ($ volume $)$ (density of rod)

$g = (S L) (d_{1}) g$

$F_{B} = ($ Volume) (density of liquid) $g$

$= (S L) (d_{2}) g$

Given that, $d_{1} < d_{2}$ . Therefore, $w < F_{B}$ Therefore, net force acting at $G$ will be

$\begin{aligned} F = F_{B} - w = (S L g) (d_{2} - d_{1}) upwards. Restoring \\ torque of this force about point P is \\ τ = F \times r_{⊥} = (S L g) (d_{2} - d_{1}) (Q G) \\ or τ = - (S L g) (d_{2} - d_{1}) \frac{L}{2} \sin θ \end{aligned}$

Here, negative sign shows the restoring nature of torque.

$or τ = - \frac{S L^{2} g (d_{2} - d_{1})}{2} θ$

$\sin θ \approx θ$ for small values of $θ$

From Eq. (i), we see that $τ \propto - θ$

Hence, motion of the rod will be simple harmonic.

Rewriting Eq. (i) as

$I \frac{d^{2} θ}{d t^{2}} = - \frac{S L^{2} g (d_{2} - d_{1})}{2} θ$

Here, $I =$ moment of inertia of rod about an axis passing through $P$ .

$I = \frac{M L^{2}}{3} = \frac{(S L d_{1}) L^{2}}{3}$

Substituting this value of $I$ in Eq. (ii), we have

$\frac{d^{2} θ}{d t^{2}} = - \frac{3}{2} \frac{g (d_{2} - d_{1})}{d_{1} L} θ$

Comparing this equation with standard differential equation of SHM, i.e. $\frac{d^{2} θ}{d t^{2}} = - ω^{2} θ$

The angular frequency of oscillation is

$ω = \sqrt{\frac{3 g (d_{2} - d_{1})}{2 d_{1} L}}$

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Simple Harmonic Motion 3 Question 11

11. A thin rod of length $L$ and uniform cross-section is pivoted at its lowest point $P$ inside a stationary homogeneous and non-viscous liquid. The rod is free to rotate in a vertical plane about a horizontal axis passing through $P$ .

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Simple Harmonic Motion 3 Question 11

11. A thin rod of length L and uniform cross-section is pivoted at its lowest point P inside a stationary homogeneous and non-viscous liquid. The rod is free to rotate in a vertical plane about a horizontal axis passing through P.

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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11. A thin rod of length $L$ and uniform cross-section is pivoted at its lowest point $P$ inside a stationary homogeneous and non-viscous liquid. The rod is free to rotate in a vertical plane about a horizontal axis passing through $P$ .