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The velocity of a particle executing a simple harmonic motion is 13ms^(-1)when its distance from the equilibrium position (Q) is 3 m and its velocity is 12 ms^(-1)when it is 5 m away from Q. The frequency of the simple harmonic motion is |
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Answer» `(5pi)/8` the VELOCITY, `v = omegasqrt(A^(2)-x^(2))` When the velocity at 3 m is 13 `ms^(-1)`, then `13^(2) = OMEGA^(2)(A^(2)-3^(2))`……(i) When the velocity at 5m is `12 ms^(-1)`, then `12^(2) = omega^(2) (A^(2)-5^(2))`........(ii) From eqns. (i) and (ii), `omega =5/4 rad s^(-1)` `u =N/(2L) = 5/(8pi)` |
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