When the length of simple pendulum is dubuled ; find the ratio of new frequency to old frequency?

Question

When the length of simple pendulum is dubuled ; find the ratio of new frequency to old frequency?

1 Answer

Impact of this question

4547 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-05T06:51:11

The answer is $= \frac{1}{\sqrt{2}}$ $=1/sqrt2$

Explanation:

The period $T$ $T$ of a simple pendulum is

$T = 2 π \sqrt{\frac{l}{g}}$ $T=2pisqrt(l/g)$

Where $l$ $l$ is the length of the pendulum

And $g$ $g$ is the acceleration due to gravity

The frequency is

$f = \frac{1}{T} = \frac{1}{2 π} \sqrt{\frac{g}{l}}$ $f=1/T=1/(2pi)sqrt(g/l)$

Squaring both sides

$f^{2} = \frac{1}{4 π^{2}} \cdot \frac{g}{l}$ $f^2=1/(4pi^2)*g/l$

Suppose l_1=2l

Then,

$f_{1}^{2} = \frac{1}{4 π^{2}} \cdot \frac{g}{2 l}$ $f_1^2=1/(4pi^2)*g/(2l)$

Then,

$\frac{f_{1}^{2}}{f^{2}} = \frac{\frac{1}{4 π^{2}} \cdot \frac{g}{2 l}}{\frac{1}{4 π^{2}} \cdot \frac{g}{l}} = \frac{1}{2}$ $f_1^2/(f^2)=(1/(4pi^2)*g/(2l))/(1/(4pi^2)*g/l)=1/2$

$\frac{f_{1}}{f} = \frac{1}{\sqrt{2}}$ $f_1/f=1/sqrt2$

Science

Math

Humanities

... and beyond

When the length of simple pendulum is dubuled ; find the ratio of new frequency to old frequency?

1 Answer

Explanation:

Related questions

Impact of this question