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# . A body executing S.H.M. along a straight line has a velocity of 3 ms–1 when it is at a distance of 4 m from its mean position and 4 ms–1 when it is at a distance of 3 m from its mean position. Its angular frequency and amplitude are

venkat
105 Points
3 years ago
$v=\omega \sqrt{a^{2}-x^{2}}$
$v_{1}=\omega \sqrt{a^{2}-x_{1}^{2}}$
$3=\omega \sqrt{a^{2}-4^{2}}$
$9=\omega ^{2}(a^{2}-16)$
$\omega ^{2}=9/(a^{2}-16)$​                -eqn(1)
$v_{2}=\omega \sqrt{a^{2}-x_{2}^{2}}$
$4=\omega \sqrt{a^{2}-{3}^{2}}$
$16=\omega^{2} ({a^{2}-9})$​                   -eqn-(2)
After simplifyng  eqns-(1) & (2) you will get,
$Angular frequency \ \omega = 1 \ rad/s$
$Amplitude \ a \ = \ 5m$
Approve my answer if it helped you.