Oscillations

JEE Main level, single correct, medium

92 of 194

Question 92 of 194

The displacement time graph of a particle executing SHM 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=3T4t=\dfrac{3T}{4}
(B) The magnitude of acceleration is maximum at t=Tt=T
(C) The speed is maximum at t=T4t=\dfrac{T}{4}
(D) The P.E.P.E. is equal to K.E.K.E. of the oscillation at t=T2t=\dfrac{T}{2}

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Answer: (A) (A),(B)\left(A\right),\left(B\right) and (C)\left(C\right)

From graph equation of SHM.
X=Acos⁡ωtX=A\cos \omega t
(A) at 3T4\dfrac{3T}{4} particle at mean position
∴a=0∴a=0
F=0F=0
(B) at T particle again at extreme position so acceleration is maximum
(C) at t=T4,t=\dfrac{T}{4}, particle is at mean position so velocity maximum. ∴a=0∴a=0
(D) KE=PEKE=PE
12k(A2−x2)=12kx2\dfrac{1}{2}k\left({A}^{2}-{x}^{2}\right)=\dfrac{1}{2}k{x}^{2}
A2=2x2{A}^{2}=2{x}^{2}
x=+A2x=\dfrac{+A}{\sqrt{2}}
A2=Acos⁡ωt\dfrac{A}{\sqrt{2}}=A\cos \omega t
t=T8t=\dfrac{T}{8}
∴A,B∴A,B and CC are correct.

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