The bob of a swinging seconds pendulum (one w | vedclass

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(C) The time period of the pendulum is $T = 2 \,s$. The angular frequency is $\omega = \frac{2 \pi}{T} = \pi \,rad/s$.At the lowest point,the displace

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$\frac{v_{0}^{2}}{4 g}$

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(C) The time period of the pendulum is $T = 2 \,s$. The angular frequency is $\omega = \frac{2 \pi}{T} = \pi \,rad/s$.At the lowest point,the displacement is $x = 0$,so the equation of motion is $x(t) = A \sin(\omega t)$,where $A$ is the amplitude.The velocity at any time $t$ is $v(t) = \frac{dx}{dt} = A \omega \cos(\omega t)$.At $t = 0$,$v(0) = v_{0} = A \omega$,so $A = \frac{v_{0}}{\omega}$.We need the height $h$ at $t = 2.25 \,s$. Since $T = 2 \,s$,$t = 2.25 \,s = T + 0.25 \,s = T + \frac{T}{8}$.At $t = \frac{T}{8} = 0.25 \,s$,the velocity is $v = v_{0} \cos(\omega \cdot \frac{T}{8}) = v_{0} \cos(\frac{2 \pi}{T} \cdot \frac{T}{8}) = v_{0} \cos(\frac{\pi}{4}) = \frac{v_{0}}{\sqrt{2}}$.Using the principle of conservation of mechanical energy: $\frac{1}{2} m v_{0}^{2} = \frac{1}{2} m v^{2} + mgh$.Substituting $v = \frac{v_{0}}{\sqrt{2}}$,we get $\frac{1}{2} v_{0}^{2} = \frac{1}{2} (\frac{v_{0}}{\sqrt{2}})^{2} + gh$.$\frac{1}{2} v_{0}^{2} = \frac{1}{4} v_{0}^{2} + gh$.$gh = \frac{1}{4} v_{0}^{2} \implies h = \frac{v_{0}^{2}}{4g}$.

The bob of a swinging seconds pendulum (one whose time period is $2 \,s$) has a small speed $v_{0}$ at its lowest point. Its height from this lowest point $2.25 \,s$ after passing through it is given by

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