Question

Time period of a simple pendulum is T. Time taken by it to complete $\dfrac{3}{8}$oscillation starting from mean position is:
A.) $\dfrac{3T}{12}$
B.) $\dfrac{3T}{8}$
C.) \[\dfrac{7T}{12}\]
D.) $\dfrac{5T}{12}$

Answer 1

Hint: The equation used in simple harmonic motion is $y=l\cos \omega t$
To complete half oscillation from extreme position, y = \[2l\]
To complete one-fourth oscillation from extreme position, y = \[l\]
To complete one oscillation from extreme position, y = \[4l\]
To complete one-eighth oscillation from extreme position, y = \[\dfrac{l}{2}\]

Complete step by step answer:
Time taken by a pendulum to complete one oscillation is called its time period.
Oscillation is a motion that repeats itself regularly.

We know for simple pendulum,

\[y=l\cos \omega t\] (1)
Where,

l = length of string
$\omega =\dfrac{2\pi }{T}$
$\dfrac{3}{8}$ Can be written as \[\dfrac{1}{4}+\dfrac{1}{8}\]

For 1 oscillation time period is T

Time taken to complete one-fourth oscillation is $\dfrac{T}{4}$ .

Time taken to complete one-eighth oscillation from extreme position is obtained from equation 1

 $y=l\cos \omega t$

\[\begin{align}
  & \dfrac{l}{2}=l\cos \dfrac{2\pi }{T}t \\
 & \dfrac{1}{2}=\cos \dfrac{2\pi }{T}t \\
 & \dfrac{\pi }{3}=\dfrac{2\pi }{T}t \\
 & t=\dfrac{T}{6} \\
\end{align}\]

Hence, total time taken by simple pendulum to complete $\dfrac{3}{8}$oscillation starting from mean position is
\[\begin{align}
  & =\dfrac{T}{4}+\dfrac{T}{6} \\
 & =\dfrac{5T}{12} \\
\end{align}\]

Note: Students should not only memorize the formula. They can forget formulas because it is difficult to learn all the formulas. So students should first learn the derivation of that formula and the basic concept of that formula before applying.