Question

if the amplitude and time period of a simple pendulum are 0.05m and 2s respectively then its maximum velocity is ?​

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

$\tt Given \begin{cases} \sf{Amplitude \: (a) \: = \: 0.05 \: m} \\ \sf{Time \: Period \: (t) \: = \: 2 \: s} \end{cases}$

We have formula for Maximum velocity

$\Large{\underline{\boxed{\sf{V_{Max} \: = \: a \: \times \: \omega}}}}$

And the formula for angular velocity is ω = 2π/t

Put value of ω

$\implies {\boxed{\sf{V_{Max} \: = \: a \: \times \: \dfrac{2 \pi}{t}}}}$

Now put values of a and t

⇒V max = 0.05 * (2 π/2)

⇒V max = 0.05 * π

⇒V max = 0.05 * 3.14

⇒V max = 0.157

$\LARGE \leadsto {\boxed{\boxed{\sf{\green{V_{Max} \: = \: 0.157 \: ms^{-1}}}}}}$

∴ Maximum Velocity is 0.157 m/s

Answer 2

Answer:

Given:

Amplitude of Pendulum is 0.05 metres

Time period = 2 seconds.

To find:

Max Velocity

Concept:

Max Velocity is obtained at the mean position of the oscillation.

Mathematically it is given as :

v max. = ω × A ,

ω is the Angular frequency and A is amplitude.

Calculation:

v max = ω × A

=> v max =( 2π/T ) × A

=> v max = (2π/2) × (0.05)

=> v max = π × (0.05)

=> v max = π/20 metres/sec.

So final answer is (π/20) m/s