Time period of a simple pendulum is T. The time taken to complete 5/8 oscillations starting from mean position is (α)/(β)T. The value of α is _________. Numerical answer type. Work it out, then open the answer.

Answer: 7. One oscillation covers 4A, so 5/8 oscillation means a path of 5/8(4A) = 5A/2. From the mean: 0 → A → 0 covers 2A in T/2. Remaining A/2 from the mean takes T/12. Total t = T/2 + T/12 = 7/12T So α = 7.

Time period of a simple pendulum is T. The time taken to complete 5/8… | JEE Main 2021 (26 Feb, Shift 2)

Q53JEE Main 2021 · 26 FEB · SHIFT 2Medium

Periodic Motion & SHM Basics

Time period of a simple pendulum is T. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{\alpha}{\beta}T$ . The value of $\alpha$ is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 7

One oscillation covers $4A$ , so $\frac{5}{8}$ oscillation means a path of $\frac{5}{8}(4A) = \frac{5A}{2}$ .
From the mean: $0 \to A \to 0$ covers $2A$ in $\frac{T}{2}$ .
Remaining $\frac{A}{2}$ from the mean takes $\frac{T}{12}$ .
Total $t = \frac{T}{2} + \frac{T}{12} = \frac{7}{12}T$
So $\alpha = 7$ .

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