First slide

Applications of SHM

Question

The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle $θ$ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

Moderate

A

$\sqrt{2 gl (1 - sinθ)}$
B

$\sqrt{2 gl (1 + cosθ)}$
C

$\sqrt{2 gl (1 - cosθ)}$
D

$\sqrt{2 gl (1 + sinθ)}$

Solution

If suppose bob rises up to a height h as shown then after releasing potential energy at extreme position becomes kinetic energy of mean position
$\Rightarrow mgh = \frac{1}{2} {mv}_{\max}^{2} \Rightarrow v_{\max} = \sqrt{2 gh}$
Also, from figure $cosθ = \frac{l - h}{l}$
$\Rightarrow h = l (1 - cosθ)$
So, $v_{\max} = \sqrt{2 gl (1 - cosθ)}$

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