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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

christy , 6 Years ago
Grade 12th pass
anser 1 Answers
venkat

Last Activity: 6 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.

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