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
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Answer:
Correct Answer: 57. (c)
Solution:
- (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 $n^{\prime}=\frac{n}{2}$
