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Two simple pendulums of lengths 10 \mathrm{~cm} and 40 \mathrm{~cm}, respectively, are given small linear displacements in one direction at the same time. They will again be in the same phase when the pendulum of shorter length has completed N_s oscillations. The value of N_s is:

Option: 1

3

 

 


Option: 2

2

 


Option: 3

4

 


Option: 4

5


Answers (1)

best_answer

The time period of a simple pendulum is given by:

T=2 \pi \sqrt{\frac{L}{g}}   where T is the time period, L is the length of the pendulum, and g is the acceleration due to gravity. 

Let's denote the time period of the shorter pendulum as T_s and the time period of the longer pendulum as T_L

For the two pendulums to be in the same phase, the shorter pendulum needs to complete an integer number of oscillations while the longer pendulum completes one oscillation.

Let's consider N_s as the number of oscillations made by the shorter pendulum for the two pendulums to be in the same phase.

Therefore, N_s \times T_s=T_L represents the condition when the two pendulums will be in the same phase.

Substituting the values, we have:

N_s \times 2 \pi \sqrt{\frac{10}{g}}=2 \pi \sqrt{\frac{40}{g}}

Canceling out the common factors of 2 \pi and  \sqrt{\frac{1}{g}} , we get: 

N_s \times \sqrt{10}=2 \times \sqrt{10}

Simplifying further, we find:

N_s=2

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