Question

$\huge{\tt{Ello!!}}$
${\underline{\underline{\textbf{Simple Pendulum:-}}}}$
❄️ A Pendulum clock normally shows correct time. On an extremely cold day, its length decreases by 0.2%. Compute the error in time per day.​

Answer 1

Answer:

Explanation:

The correct time period of pendulum clock is 2 seconds. Let L be its correct length.

∴2=2πLg−−√ …(i)

Decrease in length =0.2%=0.2100L

Length after contraction,

l=L−0.2100L=l(1−0.2100)

New time period t will be,

t=2πlg−−√=wπLg(1−0.2100)−−−−−−−−−−−√ ...(ii)

Dividing (ii) by (i) , we get

t2=(1−0.2100)1/2

or t=2(1−0.2100)1/2=2(1−12×0.2100+...)

=(2−0.2100)s

which is less than 2 seconds.

Time gained in 2 seconds=0.2100s

Total time gained in 1 day (=24×60×60s)

=0.2100×24×60×602=86.4s

Answer 2

Answer:

Given:

On a cold day, the Length of pendulum decreases by 0.2% .

To find:

Error in time period per day

Concept:

$time \: period = 2\pi \sqrt{ \dfrac{l}{g} }$

So, for very small changes in length

( < 4% ) , we can say that :

$\boxed{ \dfrac{ \Delta \: t}{t} = \dfrac{1}{2} (\dfrac{ \Delta \: l}{l} )}$

Calculation:

So time error for each oscillation

= ½ × (0.2%)

= 0.001 seconds .......(1)

Now for a whole day, we know that :

24 hr = 24 × 60 min = 24 × 60 × 60 sec

=> 24 hr = 86400 seconds.

So net error per day

$= 0.001 \times 86400 \\ = 86.4 \: seconds$

So final answer :

$\boxed{ \green{error \: per \: day = 86.4 \: secs}}$