Question

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 n oscillations. The value of n is then:
(A) 5
(B) 1
(C) 2
(D) 3

Answer 1

The time period of a simple pendulum T, depends on the length of the pendulum l and the acceleration due to gravity g. Time period T is directly proportional to the square root of the length of the pendulum l. By equating the two time periods we get the value of n.

Complete answer:
According to the question, we have two simple pendulums A and B of length ${{l}_{1}}=5$m and ${{l}_{2}}=20$m respectively, which are given small linear displacement in one direction at the same time, as shown in the diagram below.

Now, let us calculate the time period of pendulum A,
${{T}_{1}}=2\pi \sqrt{\dfrac{{{l}_{1}}}{g}}=2\pi \sqrt{\dfrac{5}{10}}=\sqrt{2}\pi$ ….(i) [ taking acceleration due to gravity $g=10$m/${{s}^{2}}$]
Then, time period of pendulum B is given by,
${{T}_{2}}=2\pi \sqrt{\dfrac{{{l}_{2}}}{g}}=2\pi \sqrt{\dfrac{20}{10}}=2\sqrt{2}\pi$ ….(ii) [ taking acceleration due to gravity $g=10$m/${{s}^{2}}$]
Now, to calculate the number of oscillations n of the pendulum A after which the two pendulums A and B will be in the phase, we compare the time period of pendulum A and time period of pendulum B calculated in equations (i) and (ii), we get,
${{T}_{2}}=2{{T}_{1}}$ ….(iii)
Therefore, we see from the above equation that the time period of pendulum A is twice the time period of pendulum B, i.e. after 2 oscillations of pendulum A, the two pendulums A and B will be in the phase.

Thus, the correct answer is option (C).

Note:
The formula of the time period of a simple pendulum should be remembered to solve such questions. The standard value of acceleration due to gravity g is $9.8$ m/${{s}^{2}}$, however we can take $g=10$m/${{s}^{2}}$to make the calculations simple and less time taking.