A simple pendulum of length 'L' is suspended from a roof of a trolley. A trolley moves in horizontal direction with an acceleration 'a'. What would be the period of
oscillation of a simple pendulum? [ \(\mathrm{g}\) is acceleration due to gravity]

\(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\)
\(\therefore\)
When the trolley has 18. (A) \(\frac{f_{1}}{24}=\frac{1.5-1}{\frac{1.5}{9 / 8}-1}=\frac{0.5}{\frac{1.5}{1.16}-1}=\frac{0.5 \times 1.16}{0.34}\)
\(\mathrm{f}_{1}=24 \times \frac{1}{2} \times \frac{1.16}{0.34}=\frac{12 \times 1.16}{0.34}=40.94 \mathrm{~cm}\)
acceleration is the resultant of \(\mathrm{a}\) and \(\mathrm{g}_{\mathrm{W}}\) where \(g\) is the effective value of acceleration due to gravity. \(\mathrm{a}\) ' in horizontal direction, the effective value \(=\)
are at right angles to each other. Hence
\(\begin{aligned} \therefore \quad T &=2 \pi \sqrt{\frac{L}{\left(a^{2}+g^{2}\right)^{\frac{1}{2}}}}=\frac{2 \pi \sqrt{L}}{\sqrt{\left(a^{2}+g^{2}\right)^{\frac{1}{2}}}} \\ &=2 \pi \sqrt{L}\left(a^{2}+g^{2}\right)^{-\frac{1}{4}} \end{aligned}\)