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This displacement time graph of a partic...

This displacement time graph of a particle executing S.H.M. is given in figure : (sketch is schematic and not to scale).

Which of the following statements is/are true for this motion?
(A) The force is zero at t = `(3T)/4`
(B)The acceleration is maximum at t=T
(C )The speed is maximum at t = `T/4`
(D) The P.E. is equal to K.E. of the oscillation at t = `T/2 `

A

(A), (B) and (D)

B

(B), (C) and (D)

C

(A), (B) and (C)

D

(A) and (D)

Text Solution

Verified by Experts

The correct Answer is:
C

F = ma ` a = - omega ^(2) x `
(A) AT `(3T)/( 4)` displacement zero ( x = 0 ), so a = 0
F = 0
(B) at t = T displacement (x) = A
x maximum , So acceleration is maximum
(C) ` v = omega sqrt(A^(2) - x^(2))`
`v_("max") at x = 0 rArr v_("max") = A omega `
At ` t = (T)/(4) , x = 0, So v_("max")`
(D) KE = PE
` :. at x = (A)/( sqrt(2))`
At, ` t = (T)/(2) x = - A ` (So not possible)
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