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In terms of time period of oscillations ...

In terms of time period of oscillations T, find the shortest time in moving a particle from `+ (A)/(2)` to `-(sqrt(3))/(2)`.

Text Solution

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The correct Answer is:
D
As shown in figure, as the SHM particle `P` moves from `P_(1)` to `P_(2)`, its corresponding particle `Q` on reference circle rotates from `Q_(1)` to `Q_(2)` an angle of `90^(@)`. From the above table , we can see that, in rotating an angle of `90^(@)` the time taken is `(T)/(4)`.
`t = T/4`
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Knowledge Check

  • Statement-I : In simple harmonic motion A is the amplitude of oscillation. If t1 be the time to reach the particle from mean position to A/sqrt2 and t_(2) the time to reach from A/sqrt2 to A . Then t_(1)=r_(2)/sqrt2 Statement-II: Equation of motion for the particle starting from mean position is given by x = ± A sinomegat and of the particle starting from extreme position is given by x = ± A cos omegat

    A
    If both statement -I and Statement -II are true , and statement-II is the correct explanation of statement -I
    B
    If both statemenrt-I and Statement-II are true but statement-II is not the correct explanation of statement -I
    C
    If statement -I is true but statement -II is false.
    D
    If statement -I is true but statement -II is true

  • The time period of an oscillator is 8 sec. the phase difference from t = 2 sec to l = 4 sec will be :-

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  • A point particle is acted upon by a restoring force –kx3 . The time period of oscillation is T when the amplitude is A. The time period for an amplitude 2A will be :

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