Question

3. Find the length of a seconds’ pendulum at a place where g= 10ms-2, π =3.14)

Answer 1

SoLuTiOn :

➡ To Find :

✏ Length of a second pendulum.

➡ Concept :

✏ We know that, Time period of second pendulum = 2s

➡ Formula :

✏ Formula of time period in terms of length of pendulum and gravitational acceleration is given by

$\star \: \underline{ \boxed{ \bold{ \rm{ \pink{T = 2\pi \sqrt{ \frac{L}{g} }}}}}} \: \star$

➡ Terms indication :

✏ T denotes time period

✏ L denotes length of pendulum

✏ g denotes gravitational acceleration

➡ Calculation :

$\mapsto \rm \: T = 2\pi \sqrt{ \frac{L}{g} } \\ \\ \rightarrow \rm \: \cancel{2} = \cancel{2 }\times 3.14 \times \sqrt{ \frac{L}{10} } \\ \\ \rightarrow \rm \: L = \frac{10}{ {(3.14)}^{2} } \\ \\ \huge\star \: \underline{ \boxed{ \bold{ \rm{ \purple{L = 1 \: m}}}}} \: \star$

➡ Additional information :

✏ Time period is defind as time requires to complete one oscillation.

✏ Unit of time period is s.

Answer 2

$\huge\bigstar{\underline{\sf{\green{Answer:-}}}}$

$\bigstar{\underline{\sf{\blue{Given:-}}}}$

• g → 10 m/s²
• π → 3.14

The time required for one oscillation by a seconds pendulum is 2 seconds.

T = 2 s

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

$\bigstar{\underline{\sf{\blue{Point\:to\: remember:-}}}}$

• l → length of the pendulum
• T → Time period
• g → acceleration due to gravity

$\bigstar{\underline{\sf{\blue{Now:-}}}}$

T² = $4 {\pi}^{2} \frac{l}{g}$

l = $\frac{ {T}^{2} g}{4 {\pi}^{2} }$

l = $\frac{ {2}^{2} × 10}{4 × {3.14}^{2}}$

l = $\frac{40}{39.43}$

$\large{\boxed{\sf{\blue{l = 1.01 m}}}}$

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