P and Q are fixed points in the same plane an | vedclass

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(C) Let the points $P$ and $Q$ be on the $x$-axis at $(-d, 0, 0)$ and $(d, 0, 0)$ respectively. The mass $m$ is at point $R$ in the $xy$-plane at $(0,

Vedclass · Accepted Answer

$2 \pi \sqrt{\frac{\sqrt{L^2-d^2}}{g}}$

Vedclass · Answer

(C) Let the points $P$ and $Q$ be on the $x$-axis at $(-d, 0, 0)$ and $(d, 0, 0)$ respectively. The mass $m$ is at point $R$ in the $xy$-plane at $(0, -h, 0)$,where $h = \sqrt{L^2 - d^2}$.When the mass is displaced slightly out of the plane (in the $z$-direction) by a distance $z$,the new distance of the mass from $P$ and $Q$ becomes $L' = \sqrt{h^2 + z^2 + d^2} = \sqrt{L^2 + z^2}$.The tension $T$ in each string is $T = \frac{mg}{2 \cos 	heta}$,where $\cos 	heta = \frac{h}{L'}$.The restoring force in the $z$-direction is $F = -2T \sin \phi$,where $\phi$ is the angle the string makes with the $xy$-plane,$\sin \phi = \frac{z}{L'}$.Thus,$F = -2 \left( \frac{mg}{2(h/L')} ight) \left( \frac{z}{L'} ight) = -\frac{mgz}{h}$.The effective spring constant is $k = \frac{mg}{h} = \frac{mg}{\sqrt{L^2 - d^2}}$.The time period is $T = 2 \pi \sqrt{\frac{m}{k}} = 2 \pi \sqrt{\frac{m}{mg/\sqrt{L^2 - d^2}}} = 2 \pi \sqrt{\frac{\sqrt{L^2 - d^2}}{g}}$.

$P$ and $Q$ are fixed points in the same plane and a mass $m$ is tied by strings as shown in the figure. If the mass is displaced slightly out of this plane and released,it will oscillate with a time period (given $PQ = 2d$,$PR = QR = L$).

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