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Physics - SHM Question with Solution | TestHub

PhysicsSHMCalculation of time periodMedium2 minPYQ_2014

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A bottle weighing $220 g$ and of area of crosssection $50 cm^{2}$ , and height $4 cm$ oscillates on the surface of water in vertical position. Its frequency of oscillation is

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Solution:

Let h be the depth of bottle in water then $A h ρ g = m g$ or $h = \frac{m}{A ρ} = \frac{200}{50 \times 1} = 4 cm$ . Now time period, $T = 2 π \frac{h}{g}$ or $v = \frac{1}{T} = \frac{1}{2 π} \frac{g}{h}$ $= \frac{7}{2 \times 22} \frac{980}{4} = 2.5 Hz$

Stream:AIIMSSubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: PYQ_2014

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