Question

If a pendulum oscillates 180 times in 3min, then find its frequency and time period?
A. 20Hz, 3sec
B. 1Hz, 18sec
C. 4Hz, 10sec
D. 1Hz, 1sec

Answer 1

As a first step convert the given time interval into its SI units. Now on recalling the definition for frequency you will realize that it is the number of oscillations taking place in one second. We have the number of oscillations taking place in 3min from which you could easily get the frequency. The reciprocal of frequency will give you the time period.

Formula used: $f=\dfrac{1}{T}$

Complete step by step answer:
In the question, we are given the number of oscillations for a pendulum in the duration of 3 minutes as 180. With this given information, we have to find the frequency as well as time period for the pendulum.
Firstly let us convert the given duration into its SI units. We know that we have 60 seconds in a minute. So,
$3\min =3\times 60\sec =180s$
Also, the number of oscillation in that duration is given as,
$n=180$
We know that frequency is the number of oscillations that take place in 1 second. So if there are 180 oscillations in 180s, then in one second the number of oscillations will be,
$180s\to 180$
$\Rightarrow 1s\to \dfrac{180}{180}=1$
Therefore, we found that only one oscillation takes place in one second. Hence the frequency will be 1Hz.
Now, we know that time period is the reciprocal of frequency and also, it is the time taken to complete one full oscillation. So, the time period T is given by,
$T=\dfrac{1}{1Hz}=1s$
Therefore, we found the time period oscillation of the given pendulum to be 1s.

So, the correct answer is “Option D”.

Note: As the options are given in SI units, make sure that you convert all the given quantities into their SI units. A simple pendulum basically consists of a mass attached to a string that is suspended from a fixed support. For a simple pendulum with wire of length L and if the acceleration due to gravity is g, then the time period of the pendulum is given by,
$T=2\pi \sqrt{\dfrac{l}{g}}$