Question

What is the length of a simple pendulum which complete half oscillation in one second

Answer 1

As per the data given in the above question.

we have to find the length of a simple pendulum which complete half oscillation in one second

you have know about the concept about simple pendulum and oscillation.

The simple pendulum is another mechanical system that moves in an oscillatory motion.

Now ,

The time period of a simple pendulum is

$t = 2\pi \: \sqrt{ \frac{l}{g} }$

where l is the length of pendulum

or

$l = \frac{g {t}^{2} }{4 {\pi}^{2} } ...........(i)$

The time period of the simple pendulum which is 2s

we have values

$t ,time = 2sec$

$g , gravitational constant= 9.8 \: \frac{m}{ {s}^{2} }$

$\pi, constant value = 3.14$

put the values in equation (i)

$l = \frac{9.8 \times ( {2}^{2} )}{4 \times ( {3.14)}^{2} }$

$l = \frac{9.8}{9.8}$

$l = 1m$

Hence

The length of a simple pendulum which complete half oscillation in one second is 1 m.

Answer 2

Answer:

The correct answer to this question is $1 m$

Explanation:

Given - The length of a simple pendulum that completes half oscillation in one second.

To Find - What is the length of a simple pendulum which completes half oscillation in one second.

Another mechanical system that operates in an oscillating motion is the simple pendulum.

According to the question,

The time period of a simple pendulum is $t = 2\pi \sqrt{\frac{l}{g}}$

and  l is the length of the pendulum,

$l = \frac{gt^{2} }{4\pi ^{2} }$ -----1)

The time period of the simple pendulum which is $2s$

The values are-

$t=time = 2 sec$

$g =$ $gravitational constant=9.8 \frac{m}{s^{2} }$

$\pi , constant value=3.14$

put the value in the equation,

$l = \frac{9.8* (2^{2} )}{4*(3.14^{2} )}$

$l = \frac{9.8}{9.8}$

$l=1m$

The length of a simple pendulum which complete half oscillation in one second is 1 m.

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