MCQ
A spring executes $SHM$ with mass of $10\,kg$ attached to it. The force constant of spring is $10\,N/m$.If at any instant its velocity is $40\,cm/sec$, the displacement will be .... $m$ (where amplitude is $0.5\,m$)
  • A
    $0.09$
  • $0.3$
  • C
    $0.03$
  • D
    $0.9$

Answer

Correct option: B.
$0.3$
b
(b) Angular velocity $\omega = \sqrt {\left( {\frac{k}{m}} \right)} $$ = \sqrt {\left( {\frac{{10}}{{10}}} \right)} = 1$
Now $u = \omega \sqrt {{a^2} - {y^2}} $

==>${y^2} = {a^2} - \frac{{{u^2}}}{{{\omega ^2}}}$$ = {(0.5)^2} - \frac{{{{(0.4)}^2}}}{{{1^2}}}$

==> ${y^2} = 0.9$

 $y = 0.3\,m$

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