Vertical displacement of a plank with a body of mass m on it is varying according to law (y= ω t+ 3 ). The minimum value of (ω) for which the mass just breaks off the plank and the moment it occurs first after t=0), are

Correct answer: A. \ g / 2 , 2 6 π g From, figure, array ll & _ R = ( 3 )^ 2 +(1)^ 2 -2 & θ= ^ -1 ( 3 1 )=(π)/(3) & y=2 (ω t+(π)/(3)) array ^ 2 y dt^ 2 =a=-2 ω^ 2 (ω t+(π)/(3) )a_ =-2 ω^ 2 =g For which mass just breaks off the plank ω= g / 2 This will be happen for the first time when array l ω t+(π)/(3)+(π)/(2) or ω t=(π)/(6) t=(π)/(6 ω)=(π)/(6) 2 g array

BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Physics
Chapter: Oscillations

BITSAT 2015 · Physics · Oscillations

Vertical displacement of a plank with a body of mass m on it is...

Vertical displacement of a plank with a body of mass

m

on it is varying according to law (y=\sin \omega t+\sqrt{3} \cos \omegat). The minimum value of (\omega) for which the mass just breaks off the plank and the moment it occurs first after

\mathrm{t}=0), are given by

Vertical displacement of a plank with a body of mass m on it is varying according to law (y= ω t+ 3 ). The minimum

Options

A\ $\sqrt{\mathrm{g} / 2}, \frac{\sqrt{2}}{6} \frac{\pi}{\sqrt{\mathrm{g}}}$ $
B( $\frac{\mathrm{g}}{\sqrt{2}}, \frac{2}{3} \sqrt{\pi / \mathrm{g}}$ )
C\ $\sqrt{\mathrm{g} / 2}, \frac{\pi}{3} \sqrt{2 / \mathrm{g}}$ $
D\$\sqrt{2 g}, \sqrt{2 \pi / 3 g}

Correct answer

A. \ $\sqrt{\mathrm{g} / 2}, \frac{\sqrt{2}}{6} \frac{\pi}{\sqrt{\mathrm{g}}}$ $

Step-by-step solution

From, figure,

\begin{array}{ll}& \Lambda_{\mathrm{R}}=\sqrt{(\sqrt{3})^{2}+(1)^{2}}-2 \\ & \theta=\tan ^{-1}\left(\frac{\sqrt{3}}{1}\right)=\frac{\pi}{3}\\ \therefore \quad & \mathrm{y}=2 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right) \end{array}

\\

\frac{\mathrmd^{2} \mathrm{y}}{\mathrm{dt}^{2}}=\mathrm{a}=-2

\omega^{2}

\sin\left(\omega \mathrm{t}+\frac{\pi}{3} \right)a_{\max }=-2 \omega^{2}=g

For which mass just breaks off the plank

\omega=\sqrt{\mathrm{g} / 2}

This will be happen for the first time when

\begin{array}{l}\omega \mathrm{t}+\frac{\pi}{3}+\frac{\pi}{2} \text{ or } \omega \mathrm{t}=\frac{\pi}{6}\\ \therefore \quad \mathrm{t}=\frac{\pi}{6 \omega}=\frac{\pi}{6} \sqrt{\frac{2}{\mathrm{ g}}} \end{array}

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