> Q. One end of massless spring of spring constant $ 100 ,N/m $ and natural length $ 0.49, m $ is fixed and other end is connected to a body of mass $ 0.5 ,kg $ lying on a frictionless horizontal table. The spring remains horizontal. If the body is made to rotate at an angular velocity of $ 2 $ rad/s, then the elongation of the spring will be – LIVE ANSWER TODAY

Q. One end of massless spring of spring constant $ 100 ,N/m $ and natural length $ 0.49, m $ is fixed and other end is connected to a body of mass $ 0.5 ,kg $ lying on a frictionless horizontal table. The spring remains horizontal. If the body is made to rotate at an angular velocity of $ 2 $ rad/s, then the elongation of the spring will be

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One end of massless spring of spring constant $ 100 ,N/m $ and natural length $ 0.49, m $ is fixed and other end is connected to a body of mass $ 0.5 ,kg $ lying on a frictionless horizontal table. The spring remains horizontal. If the body is made to rotate at an angular velocity of $ 2 $ rad/s, then the elongation of the spring will be

Solution:

Given, $M=0.5,kg$, $omega=2 rad/ s$, $l=0.49, m$, and $k=100 N/ m $

Figure represents situation, as given in question

image

Centripetal force on the blocks = Spring force

$Rightarrow Mromega^{2}=k Delta x $

$Rightarrow 0.5left(1+Delta xright)left(2right)^{2}=100cdotDelta x$

$Rightarrow left(0.49+Delta xright)4=100cdotDelta xtimes2$

$Rightarrow 0.49+Delta x=frac{200}{4} Delta x $

$Rightarrow 0.49+Delta x=50Delta x$

$49 Delta x=0.49$

$Delta x=0.01, m = 1,cm$

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