The figure shows a pendulum of length -I- suspended at a distance x vertically above a peg- The pendulum bob is deflected through an angle - and then released- When - - 90- the pendulum complete the circle- When - - 60- the velocity of the bob as it passes position E is -kl- Find the value of k g - 10 m-sec2-

When m_2 lifts up the table #160; T cosθ (90 θ)=mg T sin (90 θ)=mv^2r T sinθ =mg m_2g sinθ =mg m_2 sin(πk)=m_1 #160; ⇒sin ...

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Standard XII

Physics

Time of Flight with Ground as Frame of Reference

The figure sh...

Question

The figure shows a pendulum of length $^{'} I^{'}$ suspended at a distance $x$ vertically above a peg. The pendulum bob is deflected through an angle $θ$ and then released. When $θ = 90^{\circ}$ , the pendulum complete the circle. When $θ = 60^{\circ}$ the velocity of the bob as it passes position $E$ is $\sqrt{k l}$ . Find the value of $k (g = 10 m / s e c^{2})$ .

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Solution

When $m_{2}$ lifts up the table
$T cos θ (90 - θ) = m g$
$T sin (90 - θ) = \frac{m v^{2}}{r}$
$T sin θ = m g$
$m_{2} g sin θ = m g$
$m_{2} sin (\frac{π}{k}) = m_{1}$
$\Rightarrow sin (\frac{π}{k}) = \frac{m_{1}}{m_{2}} = \frac{3}{6} = \frac{1}{2}$
$\frac{π}{k} = \frac{π}{6}$
$\Rightarrow k = 6$

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When m2 lifts up the table Tcosθ(90−θ)=mgTsin(90−θ)=mv2rTsinθ=mgm2gsinθ=mgm2sin(πk)=m1 ⇒sin(πk)=m1m2=36=12πk=π6⇒k=6