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Two simple pendulums of length 0.5 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 x oscillations, where k is
  • a)
    1
  • b)
    3
  • c)
    2
  • d)
    5
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Two simple pendulums of length 0.5 m and 20 m respectively are given s...
Solution:

Given, the lengths of the pendulums are 0.5 m and 20 m respectively.

Let the time period of the shorter pendulum be T and the number of oscillations it completes before reaching the same phase as the longer pendulum be x.

The time period of a simple pendulum is given by the formula:

T = 2π√(l/g)

where l is the length of the pendulum and g is the acceleration due to gravity.

Calculating the time period of the shorter pendulum:

T1 = 2π√(0.5/9.8)

T1 ≈ 1.43 s

Calculating the time period of the longer pendulum:

T2 = 2π√(20/9.8)

T2 ≈ 12.65 s

The time taken for the shorter pendulum to complete x oscillations is given by:

t = xT1

The time taken for the longer pendulum to complete x oscillations is given by:

t = xT2

For the two pendulums to be in phase, the time taken for both of them to complete x oscillations should be the same.

Therefore,

xT1 = xT2

x(2π√(0.5/9.8)) = x(2π√(20/9.8))

Simplifying the equation, we get:

x = 2

Therefore, the pendulum of shorter length will complete 2 oscillations when both pendulums are in phase again.

Hence, the correct answer is option C.
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Community Answer
Two simple pendulums of length 0.5 m and 20 m respectively are given s...
I think the question is wrong, it's 2m not 20m. if taken as 2, the correct answer is C
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