Q133 / 150 · MHT CET 2020 (12 Oct Shift 2) · Physics
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]
- Correct answerA$2 \pi \sqrt{L}\left(a^{2}+g^{2}\right)^{-\frac{1}{4}}$
- B$2 \pi \sqrt{L}\left(a^{2}+g^{2}\right)^{-\frac{1}{2}}$
- C$2 \pi \sqrt{\frac{L}{g+a}}$
- D$2 \pi \sqrt{\frac{L}{g-a}}$
Solution
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$
$\therefore$
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 $\begin{aligned} \therefore \quad T &=2 \pi \sqrt{\frac{L}{\left(a^{2}+g^{2}\right)^{\frac{1}{2}}}}=\frac{2 \pi \sqrt{L}}{\sqrt{\left(a^{2}+g^{2}\right)^{\frac{1}{2}}}} \\ &=2 \pi \sqrt{L}\left(a^{2}+g^{2}\right)^{-\frac{1}{4}} \end{aligned}$
Hence, the correct answer is option: 'A'.