An object performs `S.H.M.` of amplitude `5 cm` and time period `4 s` at `t = 0` the object is at the centre of the oscillation i.e., `x = 0` then calculate. (i) Frequency of oscillation , (ii) The displacement at `0.5 s` (iii) The maximum acceleration of the object ,
(iv) The velocity at a displacement of `3 cm`.
Text Solution
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(i) Frequency `f = (1)/(T) = (1)/(4) = 0.25 Hz` (ii) The displacement equation of object `x = Asinomegat` at `t = 0.5` s, `x = 5sin(2pi xx 0.25 xx 0.5) = 5sin"(pi)/(4) = (pi)/(sqrt(2)) cm` (iii) Maximum accleration `a_("max") = omega^(2)A = (0.5 pi)^(2) xx 5 = 12.3 cm//s^(2)` (iv) Velocity at `x = 3 cm` is `v = +- omegasqrt(A^(2) - x^(2)) = +- 0.5pisqrt(5^(2)-3^(2)) = +- 6.28 cm//s`
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