Question

Find the length of a seconds' pendulum at a place where g=10m/s² . Take( π = 3.14)

Answer 1

Answer:

1 m

Explanation:

The second's pendulum is a special type of pendulum which has a Time period (T) of 2 seconds.

The formula for calculating Time Period is given as:

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

We know that,

• T = 2 seconds
• g = 10 m/s²
• π = 3.14

Substituting the values we get:

$\implies 2 = 2 \times 3.14 \times \sqrt{\dfrac{l}{10}}\\\\\implies 2 = 6.28 \times \sqrt{\dfrac{l}{10}}\\\\\\\text{Squaring on both sides we get:}\\\\\implies 4 = 39.44 \times \dfrac{l}{10}\\\\\implies 4 = 3.94 \times l\\\\\boxed{\implies l = \dfrac{4}{3.94} \approx 1 m}$

Hence the length of a second's pendulum is approximately 1 m.

Answer 2

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

where ,

• T denotes Time Interval
• l denotes length of the pendulum
• g denotes gravity

Given ,

We need to find " the length of a seconds' pendulum at a place

where g = 10m/s² "

Time Interval , T = 2 sec [ Since given second's pendulum ]

Gravity , g = 10 m/s²

Length of the pendulum , l = ?

$\implies \bf T=2\pi\sqrt{\dfrac{l}{g}} \\\\\implies \bf T^2=4\pi^2\dfrac{l}{g}[\;Squaring\ on\ both\ sides\ ]\\\\\implies \bf l=\dfrac{T^2g}{4\pi^2}\\\\\implies \bf l=\dfrac{4*10}{4*(3.14)^2}\\\\\implies \bf l=1.01\ m$

So the length of the second's pendulum is 1.01 meters