A bob of a simple pendulum has mass 'm' and is oscillating with an amplitude 'a'. If the length of the…

A bob of a simple pendulum has mass 'm' and is oscillating with an amplitude 'a'. If the length of the pendulum is 'L' then the maximum tension in the string is $\left[\cos 0^{\circ}=1, \mathrm{~g}=\right.$ acceleration due to gravity $]$
  1. $\mathrm{mg}\left[1+\left(\frac{\mathrm{a}}{\mathrm{L}}\right)^{2}\right]$
  2. $\mathrm{mg}\left[1-\left(\frac{\mathrm{L}}{\mathrm{a}}\right)^{2}\right]$
  3. $\mathrm{mg}\left[1+\left(\frac{\mathrm{L}}{\mathrm{a}}\right)^{2}\right]$
  4. $\mathrm{mg}\left[1-\left(\frac{\mathrm{a}}{\mathrm{L}}\right)^{2}\right]$

Solution

tension in the string is maximum when the bob passes through the mean position. $\mathrm{T}_{\max }=\mathrm{mg}+\frac{\mathrm{mV}^{2}}{\mathrm{~L}}$...(1) In S.H.M. velocity at the mean position is given by $\mathrm{V}=\mathrm{a} \omega$ For simple pendulum $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ $\therefore \omega=\frac{2 \pi}{T}=\sqrt{\frac{g}{L}}$ $\therefore V=a \sqrt{\frac{g}{L}}$ or $V^{2}=a^{2} \frac{g}{L}$ Putting this value of $\mathrm{V}^{2}$ in $\mathrm{Eq} .(1)$ we get $\mathrm{T}_{\max }=\mathrm{mg}\left[1+\left(\frac{\mathrm{a}}{\mathrm{L}}\right)^{2}\right]$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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