Oscillations - Result Question 53

57. Two simple pendulums of length 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

[1998]

(a) 5

(b) 1

(c) 2

(d) 3

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Answer:

Correct Answer: 57. (c)

Solution:

  1. (c) Let the pendulums be in phase after tsec 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 =2π20g×n

& the time of (n+1) oscillation

=2π5g×(n+1)

To be in phase

2π20g×n=2π5g×(n+1)

or, 2n=n+1

or, n=1

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

58 .

(c) n=12πkm

n=12πk4m=12×12πkm

On putting the value of n we get n=n2



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