Two masses,both equal to 100, g,are suspended | vedclass

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(A) The period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.Given $\ell = 1.0\, m$ and taking $g = 10\, m/s^2$,we have $T = 2\pi

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at $	heta = 0.0^o$,$0.50\, s$ later.

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(A) The period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.Given $\ell = 1.0\, m$ and taking $g = 10\, m/s^2$,we have $T = 2\pi \sqrt{\frac{1}{10}} \approx 2\pi 	imes 0.316 \approx 1.987\, s \approx 2.0\, s$.Since the period $T$ is independent of the amplitude for small angles,both pendulums have the same period $T = 2.0\, s$.They are released from rest at $t = 0$. The time taken for a pendulum to reach the vertical position $(	heta = 0^o)$ from its extreme position is one-quarter of its time period,i.e.,$t = \frac{T}{4}$.Thus,$t = \frac{2.0}{4} = 0.5\, s$.Both pendulums reach the vertical position at the same time $t = 0.5\, s$,so they will collide at $	heta = 0.0^o$ after $0.50\, s$.

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