Question

a one metre long simple pendulum comoletes 15 vibrations in 30 seconds. calculate the value of 'g' at that place​

Answer 1

Given:

A one metre long simple pendulum comoletes 15 vibrations in 30 seconds.

To find:

Value of gravitational acceleration at that place.

Calculation:

Time period of a simple pendulum is defined as the time taken to complete one full oscillation.

So, let time period be t ;

$\therefore \: \rm{t = \dfrac{total \: time}{total \: vibrations} = \dfrac{30}{15} = 2 \: sec}$

Now , general expression of time period for a simple pendulum is given as follows:

$\rm{ \therefore \: t = 2\pi \sqrt{ \dfrac{L}{g} } }$

$\rm{ = > \: 2= 2\pi \sqrt{ \dfrac{1}{g} } }$

$\rm{ = > \: 1= \pi \sqrt{ \dfrac{1}{g} } }$

$\rm{ = > \: \sqrt{ \dfrac{1}{g} } = \dfrac{1}{\pi} }$

$\rm{ = > \: \dfrac{1}{g} = \dfrac{1}{ {\pi}^{2} } }$

$\rm{ = > \: g = {\pi}^{2} \: m {s}^{ - 2} }$

So, final answer is:

$\boxed{ \sf{ \: g = {\pi}^{2} \: m {s}^{ - 2} }}$

Answer 2

Question:-

An one metre long simple pendulum comoletes 15 vibrations in 30 seconds. Calculate the value of acceleration due to gravity at that place.

Answer:-

Time period is defined as the time taken to complete one full oscillation by a pendulum.

$\sf{\\}$

It is given that,

$\scriptsize{\textsf{Time taken to complete 15 oscillations = 30 seconds}}$

$\longrightarrow \scriptsize{\textsf{Time taken to complete 1 oscillation = 2 seconds}}$

Hence, Time period = 2 seconds

$\sf{\\}$

Let the Acceleration due to gravity at that place be g'

We know that

$\sf{T = 2 \pi \sqrt{\dfrac{l}{g'}}}$

$\longrightarrow \sf{2 = 2 \pi \sqrt{\dfrac{1}{g'}}}$

$\longrightarrow \sf{1 = 1 \pi \sqrt{\dfrac{1}{g'}}}$

$\longrightarrow \sf{{1}^{2} = {\Bigg\lgroup1 \pi \sqrt{\dfrac{1}{g'}}\Bigg\rgroup}^{2}}$

$\longrightarrow \sf{1 = {\pi}^{2} \dfrac{1}{g'}}$

$\longrightarrow \sf{g' = {\pi}^{2}\:m/{s}^{2}}$

Hence,

$\boxed{\red{\scriptsize{\sf{Ans. \: g' = Acceleration\: due\: to \:gravity\: at\:that\:place = {\pi}^{2} \: m/{s}^{2}}}}}$