Oscillations - Result Question 53

57. Two simple pendulums of length $5 m$ and $20 m$ 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

[1998]

(a) 5

(b) 1

(c) 2

(d) 3

Show Answer

Answer:

Correct Answer: 57. (c)

Solution:

  1. (c) Let the pendulums be in phase after $\boldsymbol{{}t} sec$ of start. Within this time, if the bigger pendulum executes $n$ oscillations, the smaller one will have executed $(n+1)$ oscillations.

Now, the time of $n$ oscillation $=2 \pi \sqrt{\frac{20}{g}} \times n$

& the time of $(n+1)$ oscillation

$=2 \pi \sqrt{\frac{5}{g}} \times(n+1)$

To be in phase

$2 \pi \sqrt{\frac{20}{g}} \times n=2 \pi \sqrt{\frac{5}{g}} \times(n+1)$

or, $2 n=n+1$

or, $n=1$

Hence, the no. of oscillations executed by shorter pendulum $=n+1=1+1=2$

58 .

(c) $n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$

$n^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{k}{4 m}}=\frac{1}{2} \times \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$

On putting the value of $n$ we get $n^{\prime}=\frac{n}{2}$