
A simple pendulum completes 40 oscillations in a minute.Find its (a) frequency (b) time period.
Answer
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Hint:In order to calculate the frequency and time period of the pendulum which completes 40 oscillations in a minute, we need to find it using the pendulum formula. We will find it one by one.
Complete step-by-step answer:
From the question, it is given that,
Number of Oscillations, \[N = 40\]
And Time\[~t=1min\]
Now,
(a) Frequency: It is given by oscillations per second.
So,
Frequency \[f = \dfrac{N}{t}\]
$ = \left( {\dfrac{{oscillations}}{{60}}} \right){s^{ - 1}}$
$ = \left( {\dfrac{{40}}{{60}}} \right){s^{ - 1}}$
$ = 0.667Hz$
(b) Time period: It is given by 1 upon frequency.
So,
$\left( {\dfrac{1}{{freq}}} \right)$
$ = \left( {\dfrac{1}{{0.67}}} \right)s$
$ = 1.5s$
Hence, frequency is $0.667Hz$ and time period is $1.5s$.
Additional Information: Simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position. Also, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. The time interval of each complete vibration is the same.Whilst simple harmonic motion is a simplification, it is still a very good approximation. Simple harmonic motion is important in research to model oscillations for example in wind turbines and vibrations in car suspensions.
Note:While solving this question, we should know that frequency of a pendulum is oscillations of it per second and the time period is given by $\dfrac{1}{{freq}}$. We have to use this formula and put the values of the variables provided in the question carefully in order to get the answer.
Complete step-by-step answer:
From the question, it is given that,
Number of Oscillations, \[N = 40\]
And Time\[~t=1min\]
Now,
(a) Frequency: It is given by oscillations per second.
So,
Frequency \[f = \dfrac{N}{t}\]
$ = \left( {\dfrac{{oscillations}}{{60}}} \right){s^{ - 1}}$
$ = \left( {\dfrac{{40}}{{60}}} \right){s^{ - 1}}$
$ = 0.667Hz$
(b) Time period: It is given by 1 upon frequency.
So,
$\left( {\dfrac{1}{{freq}}} \right)$
$ = \left( {\dfrac{1}{{0.67}}} \right)s$
$ = 1.5s$
Hence, frequency is $0.667Hz$ and time period is $1.5s$.
Additional Information: Simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position. Also, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. The time interval of each complete vibration is the same.Whilst simple harmonic motion is a simplification, it is still a very good approximation. Simple harmonic motion is important in research to model oscillations for example in wind turbines and vibrations in car suspensions.
Note:While solving this question, we should know that frequency of a pendulum is oscillations of it per second and the time period is given by $\dfrac{1}{{freq}}$. We have to use this formula and put the values of the variables provided in the question carefully in order to get the answer.
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