Question

The time period of oscillations of a simple pendulum
is 1 minute. If its length is increased by 44%. then
its new time period of oscillation will be
​

Answer 1

Answer:

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Explanation:

Time period of a simple pendulum at surface of earth is given by [T=2π√(l/g)] where g=acceleration due to gravity so now if the length is increased by 44%. New length= 1.44l

Answer 2

The new times period is $\dfrac{12}{10}$ times the original time period .

Let , length be l .

Therefore , time period t :

$T=2\pi\sqrt{\dfrac{l}{g}}$

Now, when length is increased by 44%.

So , new length ,

$l'=l+\dfrac{44}{100}l\\\\l'=\dfrac{144}{100}l$

Therefore , new time period ,

$T'=2\pi \sqrt{\dfrac{l'}{g}}\\\\T'=2\pi \sqrt{\dfrac{\dfrac{144}{100}l}{g}}\\\\T'=\dfrac{12}{10}(2\pi\sqrt{\dfrac{l}{g}})\\\\T'=\dfrac{12}{10}T$

Therefore the new times period is $\dfrac{12}{10}$ times the original time period .

Hence , this is the required solution .

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