The path length of oscillation of simple pendulum of length 1m is 16cm . Its maximum velocity is (g = (π )2 (m s)-2)

Question

The path length of oscillation of simple pendulum of length 1m is 16cm . Its maximum velocity is (g = (π )2 (m s)-2) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-zvwxnczr8l4uvbt0.jpg" /> (A) 2π cms- 1 (B) 4π cms- 1 (C) 8π cm s- 1 (D) 16π cms- 1

Tardigrade Admin · Accepted Answer

Path length = 1 6 text cm ∴ Amplitude A = 8 text cm <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-zvwxnczr8l4uvqcr.jpg" /> Period T = 2 π √(l/g) = 2 π √(l/π 2) = 2 s = 2 π × (1/π ) = 2 s Maximum velocity V textmax = = a ω = A × (2 π /Î¤) = 8 × (2 π /2) = 8 π c m s- 1

Q. The path length of oscillation of simple pendulum of length $1 m$ is $16 c m$ . Its maximum velocity is $(g = (π)^{2} (m s)^{- 2})$

$2 π c m s^{- 1}$

$4 π c m s^{- 1}$

$8 π c m s^{- 1}$

$16 π c m s^{- 1}$

Path length $= 16 cm$
$∴$ Amplitude $A = 8 cm$

Period $T = 2 π \frac{l}{g}$
$= 2 π \frac{l}{π ^{2}} = 2 s$
$= 2 π \times \frac{1}{π} = 2 s$
Maximum velocity $V_{max =} = aω$
$= A \times \frac{2 π}{T}$
$= 8 \times \frac{2 π}{2}$
$= 8 π c m s^{- 1}$

Q. The path length of oscillation of simple pendulum of length 1m is 16cm . Its maximum velocity is (g=(π)2(ms)−2)

2πcms−1

4πcms−1

8πcms−1

16πcms−1