Question

Two simple pendulums of length $0.5m$ and $20m$ respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed oscillations $n{T}_1=(n-1)T_2$, where $T_1$ is time period of shorter length & $T_2$ be time period of longer wavelength and $n$ are no. of oscillations completed. Find the value of $n$.

• A. $5$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

Let $T_1$, $T_2$ be the time period of shorter length and longer length pendulums respectively. As per question, $n{T}_1=(n-1)T_2$
So, $n2\pi \sqrt{\frac{0.5}{g}}=(n-1)2\pi \sqrt{\frac{20}{g}}$
Or $n=(n-1)\sqrt{40}\approx (n-1)6$
Hence $n=6/5\approx 1$