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(A) Let the original length of the spring be $l$ and the spring constant be $k$. When the body is whirled in a horizontal circle,the centripetal force

Vedclass · Accepted Answer

$15$

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(A) Let the original length of the spring be $l$ and the spring constant be $k$. When the body is whirled in a horizontal circle,the centripetal force is provided by the spring force $F = kx$,where $x$ is the elongation.The radius of the circular path is $r = l + x$.The centripetal force is $F = m r \omega^2 = m(l + x) \omega^2$.Case $1$: Elongation $x_1 = 1\, cm$,angular velocity $\omega_1 = \omega$.$k(1) = m(l + 1) \omega^2$  --- $(i)$Case $2$: Elongation $x_2 = 5\, cm$,angular velocity $\omega_2 = 2\omega$.$k(5) = m(l + 5) (2\omega)^2 = 4m(l + 5) \omega^2$ --- $(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{5k}{k} = \frac{4m(l + 5) \omega^2}{m(l + 1) \omega^2}$$5 = \frac{4(l + 5)}{l + 1}$$5(l + 1) = 4(l + 5)$$5l + 5 = 4l + 20$$l = 15\, cm$.

$A$ body of mass $m$ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is $1\, cm$. If the angular velocity is doubled,the elongation in the spring is $5\, cm$. The original length of the spring is ......... $cm$

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