Time period T of a simple pendulum of length l is given by T -2 - l - g - If the length is increased by 2 - then an approximate change in the time period isA- 2 -B- 1 -C- 1-2-D- None of these

The time period of simple pendulum is T. When the length increases by 10 cm, its period is $T_{1}$ . When the length is decreased by 10 cm, its period is $T_{2}$ .Then the relation between T, $T_{1}$ and $T_{2}$ is

Q. The time

t

, of a complete oscillation of a simple pendulum of length

l

is given by

t = 2 π \sqrt{\frac{l}{g}}

where

g

is gravitational constant. Find the approximate percentage of errors in

t

when the percentage of error

l

1

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Time period T of a simple pendulum of length l is given by T=2π√lg. If the length is increased by 2%, then an approximate change in the time period is

The correct option is B 1%dTdl=π√gl△T=dTdl△l=π√gl.2l100=T100⇒△TT=0.01 Approximate change =1%

Time period $T$ of a simple pendulum of length $l$ is given by
$T = 2 π \sqrt{\frac{l}{g}} .$ If the length is increased by $2 %$ , then an approximate change in the time period is

The correct option is B $1$ %
$\frac{d T}{d l} = \frac{π}{\sqrt{g l}} △ T = \frac{d T}{d l} △ l = \frac{π}{\sqrt{g l}} . \frac{2 l}{100} = \frac{T}{100} \Rightarrow \frac{△ T}{T} = 0.01$
Approximate change $= 1$ %