Question

If the time period of a two-meter-long simple pendulum is 2 s, the acceleration due to gravity at the place where the pendulum is executing S.H.M. is:
(a) $2{\pi ^2}{\rm{ m}}{{\rm{s}}^{ - 2}}$
(b) $16{\rm{ m}}{{\rm{s}}^{ - 2}}$
(c) $9.8{\rm{ m}}{{\rm{s}}^{ - 2}}$
(d) ${\pi ^2}{\rm{ m}}{{\rm{s}}^{ - 2}}$

Answer 1

Hint: We will use the time period formula for simple pendulum to solve this question. Using the formula of time period of pendulum we can easily find the value of “g”.

Formula used:
${\rm{T}} = 2\pi \sqrt {\dfrac{l}{g}} $
Where,
T is the known as the time period of the pendulum.
l is the length of the pendulum and
g is the acceleration due to gravity.

Complete answer:
According to the question it is given that the length of the pendulum is 2 meters and the time period of the given pendulum is 2 s. And to calculate the acceleration due to gravity simply substitute these values in the above formula.
\[{\rm{T}} = 2\pi \sqrt {\dfrac{l}{g}} \]
\[ \Rightarrow 2 = 2\pi \sqrt {\dfrac{2}{g}} \]
\[ \Rightarrow g = 2{\pi ^2}\]

Hence the acceleration due to the gravity of the pendulum is $2{\pi ^2}{\rm{ m}}{{\rm{s}}^{ - 2}}$ . Option (a) is correct.

Note: The net acceleration given to objects as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity or g. A vector quantity, that is. A light, inextensible thread with an upper end fastened to sturdy support is used to hang a point mass M. The mass is moved away from its centre of gravity.