Question

find the percentage change in the time period of simple pendulum if it is taken to a height half the radius of Earth​

Answer 1

Answer:

Given:

Pendulum is taken to height half of the radius of Earth

To find:

% Change in time period of pendulum

Calculation:

First Calculate the new gravitational acceleration :

height be h , radius be r , g be gravity at surface

$g2 = \dfrac{g}{ {(1 + \frac{h}{r}) }^{2} }$

Putting h = r/2

$= > g2 = \dfrac{g}{ {(1 + \frac{1}{2} )}^{2} }$

$= > g2 = \dfrac{4g}{9}$

Now, Initial time period of Pendulum :

$t1 = 2\pi \sqrt{ \dfrac{l}{g} }$

Final time period:

$t2 = 2\pi \sqrt{ \dfrac{l}{( \frac{4g}{9}) } } \\ = > t2 = 2\pi \sqrt{ \frac{9l}{4g} }$

$= > t2 = \dfrac{3}{2} \times t1$

So change in time period :

$= t2 - t1 \\ = \frac{3}{2} t1 - t1 \\ = ( \frac{1}{2}) t1$

So percentage change :

$= \dfrac{ \Delta \: t}{t1} \times 100\% \\ = \dfrac{1}{2} \times 100\% \\ = 50\%$

So there is 50% increase in Time period.

Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

Pendulum is taken to height half of the radius of Earth

To find:

% Change in time period of pendulum

Calculation:

First Calculate the new gravitational acceleration :

height be h , radius be r , g be gravity at surface

g2 = \dfrac{g}{ {(1 + \frac{h}{r}) }^{2} }

Putting h = r/2

= > g2 = \dfrac{g}{ {(1 + \frac{1}{2} )}^{2} }

= > g2 = \dfrac{4g}{9}

Now, Initial time period of Pendulum :

t1 = 2\pi \sqrt{ \dfrac{l}{g} }

Final time period:

t2 = 2\pi \sqrt{ \dfrac{l}{( \frac{4g}{9}) } } \\ = > t2 = 2\pi \sqrt{ \frac{9l}{4g} }

= > t2 = \dfrac{3}{2} \times t1

So change in time period :

= t2 - t1 \\ = \frac{3}{2} t1 - t1 \\ = ( \frac{1}{2}) t1

So percentage change :

= \dfrac{ \Delta \: t}{t1} \times 100\% \\ = \dfrac{1}{2} \times 100\% \\ = 50\%

So there is 50% increase in Time period.