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4-1.Newton's Laws of Motion
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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 one centimetre. If the angular velocity is doubled, the elongation in the spring is $5\, cm$ . The original length of the spring is ............ $cm$
A$16$
B$15$
C$14$
D$13$
Solution
Let $'l'$ be the original length.
Given ${\rm{m}}{\omega ^2}\left( {l + {{\rm{x}}_1}} \right) = {\rm{k}}{{\rm{x}}_1}$
$m{(2\omega )^2}\left( {l + {x_2}} \right) = k{x_2}$
where $x_{1}=1 \mathrm{cm} \quad \mathrm{x}_{2}=5 \mathrm{cm}$
$\therefore \frac{{4\left( {l + {{\rm{x}}_2}} \right)}}{{l + {{\rm{x}}_1}}} = \frac{{{{\rm{x}}_2}}}{{{{\rm{x}}_1}}}$
$4{x_1}\left( {l + {x_2}} \right) = {x_2}\left( {l + {x_1}} \right)$
$4(l + 5) = 5(l + 1)$
$l = 15{\rm{cm}}$
Given ${\rm{m}}{\omega ^2}\left( {l + {{\rm{x}}_1}} \right) = {\rm{k}}{{\rm{x}}_1}$
$m{(2\omega )^2}\left( {l + {x_2}} \right) = k{x_2}$
where $x_{1}=1 \mathrm{cm} \quad \mathrm{x}_{2}=5 \mathrm{cm}$
$\therefore \frac{{4\left( {l + {{\rm{x}}_2}} \right)}}{{l + {{\rm{x}}_1}}} = \frac{{{{\rm{x}}_2}}}{{{{\rm{x}}_1}}}$
$4{x_1}\left( {l + {x_2}} \right) = {x_2}\left( {l + {x_1}} \right)$
$4(l + 5) = 5(l + 1)$
$l = 15{\rm{cm}}$
Standard 11
Physics
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