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Q13.17: NCERT Physics - Part 2 OSCILLATIONS 13.17 Solution: A cylindrical piece of cork of density of base area A and…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently write the restoring force as $-( ho- ho_l)gA x$, mixing the equilibrium weight balance with the small‑displacement term; the correct linear restoring force comes solely from the extra displaced liquid, $F_{rest}= -\rho_l g A x$, because the weight of the cork and the buoyant force at equilibrium already cancel.

A cylindrical piece of cork of density of base area A and height h floats in a liquid of density _l. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period T = 2��(h_/_l g) where _ is the density of cork. (Ignore damping due to viscosity of the liquid).

Step-by-Step Verified Solution

**Answer:** T=2πhρρlgT = 2\pi\sqrt{\frac{h\rho}{\rho_l g}} **Step-by-Step Explanation:** In equilibrium, weight of cork = weight of displaced liquid. Let the cork be depressed by x. Extra upthrust = restoring force:
F=−(Volume displaced)ρlg=−(Ax)ρlgF = -(\text{Volume displaced}) \rho_l g = -(Ax)\rho_l g
Acceleration: a=F/ma = F/m where m=Ahρm = Ah\rho.
a=−AxρlgAhρ=−ρlghρxa = -\frac{Ax\rho_l g}{Ah\rho} = -\frac{\rho_l g}{h\rho}x
Since a∝−xa \propto -x, motion is SHM with:
ω2=ρlghρ⇒T=2πhρρlg\omega^2 = \frac{\rho_l g}{h\rho} \Rightarrow T = 2\pi\sqrt{\frac{h\rho}{\rho_l g}}
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