A pendulum of length L swings from rest to rest n times in one second. The value of acceleration due to gravity is

Question

A pendulum of length L swings from rest to rest n times in one second. The value of acceleration due to gravity is (A) 4π2n2L (B) 2π2n2L (C) π2n2L (D) (π2n2L/2)

Tardigrade Admin · Accepted Answer

T =2 π √( l / g ) frequency =(n/2) ⇒ T=(2/n), As its given rest to rest as n . (2/n)=2 π √(1/g) ∴ g=n2 π2 L

Q. A pendulum of length L swings from rest to rest n times in one second. The value of acceleration due to gravity is

$4 π^{2} n^{2} L$

$2 π^{2} n^{2} L$

$π^{2} n^{2} L$

$\frac{π ^{2} n ^{2} L}{2}$

$T = 2 π \frac{l}{g}$
frequency $= \frac{n}{2} \Rightarrow T = \frac{2}{n}$ ,
As its given rest to rest as $n .$
$\frac{2}{n} = 2 π \frac{1}{g}$
$∴ g = n^{2} π^{2} L$

Q. A pendulum of length L swings from rest to rest n times in one second. The value of acceleration due to gravity is

4π2n2L

2π2n2L

π2n2L

2π2n2L​

T=2πgl​​ frequency =2n​⇒T=n2​, As its given rest to rest as n. n2​=2πg1​​ ∴g=n2π2L

$4 π^{2} n^{2} L$

$2 π^{2} n^{2} L$

$π^{2} n^{2} L$

$\frac{π ^{2} n ^{2} L}{2}$

$T = 2 π \frac{l}{g}$
frequency $= \frac{n}{2} \Rightarrow T = \frac{2}{n}$ ,
As its given rest to rest as $n .$
$\frac{2}{n} = 2 π \frac{1}{g}$
$∴ g = n^{2} π^{2} L$