A particle of mass m attached with a massless spring natural length l and stiffness k is released from rest from the horizontal position of the relaxed spring. When the particle passes through its lowest point, the maximum length of the spring becomes 2l. Find the:
a. speed of the particle at its lowest point.
b. acceleration of the particle at its lowest position.

\sqrt{2 g l - \frac{k}{m} l^{2}}

\frac{K l - m g}{m}

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\sqrt{14 g l - \frac{k}{m} l^{2}}

\frac{K l - m g}{m}

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\sqrt{4 g l - \frac{k}{m} l^{2}}

\frac{K l - m g}{m}

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\sqrt{3 g l - \frac{k}{m} l^{2}}

\frac{K l - 2 m g}{m}

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Solution

The correct option is C a. $\sqrt{4 g l - \frac{k}{m} l^{2}}$
b. $\frac{K l - m g}{m}$

$a) Loss in gravititational P.E= gain in spring E$

$\Rightarrow mg \times 2 l = \frac{1}{2} k (2 l - 1)^{2} + \frac{1}{2} m v^{2}$
$\Rightarrow 2 m g l = \frac{1}{2} k l^{2} + \frac{1}{2} m v^{2}$
$\Rightarrow \frac{1}{2} m v^{2} = 2 m g l - \frac{1}{2} k l^{2}$
$\Rightarrow v^{2} = 4 g l - \frac{k}{m} l^{2}$

$[v = \sqrt{4 g l - \frac{k}{m} l^{2}}] A n s$

$b)$
$k l - m g = m a$
$\Rightarrow \frac{k l - m g}{m} = a$
$∴ [a = \frac{k l - m g}{m}] Ans$

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The correct option is C a. √4gl−kml2b. Kl−mgm a) Loss in gravititational P.E= gain in spring E ⇒mg×2l=12k(2l−1)2+12mv2⇒2mgl=12kl2+12mv2⇒12mv2=2mgl−12kl2⇒v2=4gl−kml2 [v=√4gl−kml2]Ans b) kl−mg=ma⇒kl−mgm=a∴[a=kl−mgm] Ans