JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 12)
The displacement time graph of a particle
executing S.H.M is given in figure :
(sketch is schematic and not to scale)_2nd_September_Evening_Slot_en_12_1.png)
Which of the following statements is/are true for this motion?
(A) The force is zero at t = $${{3T} \over 4}$$
(B) The acceleration is maximum at t = T
(C) The speed is maximum at t = $${{T} \over 4}$$
(D) The P.E. is equal to K.E. of the oscillation at t = $${{T} \over 2}$$
(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 = $${{3T} \over 4}$$
(B) The acceleration is maximum at t = T
(C) The speed is maximum at t = $${{T} \over 4}$$
(D) The P.E. is equal to K.E. of the oscillation at t = $${{T} \over 2}$$
(B), (C) and (D)
(A), (B) and (C)
(A) and (D)
(A), (B) and (D)
Explanation
(A) F = ma and a = $$-\omega^{2} x$$
At $$\frac{3T}{4} $$ displacement zero (x = 0), so a = 0
$$ \therefore $$ F = 0
(B) at t = T, Particle is at extreme.
displacement (x) = A $$ \Rightarrow $$ x maximum,
So acceleration is maximum.
(C) V = $$\omega \sqrt{A^{2}-x^{2}} $$
at t = $$\frac{T}{4} $$, x = 0
$$ \therefore $$ V = A$$\omega $$ = Vmax
(D) KE = PE
$$\frac{1}{2} k\left( A^{2}-x^{2}\right) =\frac{1}{2} kx^{2}$$
$$ \Rightarrow $$ A2 – x2 = x2
$$ \Rightarrow $$ A2 = 2x2
A = $$\sqrt{2} $$x
$$ \Rightarrow $$ x = $$\frac{A}{\sqrt{2} } $$ = A cos $$\omega $$t
$$ \Rightarrow $$ cos $$\omega $$t = $$\frac{1}{\sqrt{2} } $$
$$ \Rightarrow $$ $$\omega $$t = $$\frac{\pi }{4} $$
$$ \Rightarrow $$ $$\frac{2\pi }{T} \cdot t=\frac{\pi }{4} $$
$$ \Rightarrow $$ t = $$\frac{T}{8} $$
At $$\frac{3T}{4} $$ displacement zero (x = 0), so a = 0
$$ \therefore $$ F = 0
(B) at t = T, Particle is at extreme.
displacement (x) = A $$ \Rightarrow $$ x maximum,
So acceleration is maximum.
(C) V = $$\omega \sqrt{A^{2}-x^{2}} $$
at t = $$\frac{T}{4} $$, x = 0
$$ \therefore $$ V = A$$\omega $$ = Vmax
(D) KE = PE
$$\frac{1}{2} k\left( A^{2}-x^{2}\right) =\frac{1}{2} kx^{2}$$
$$ \Rightarrow $$ A2 – x2 = x2
$$ \Rightarrow $$ A2 = 2x2
A = $$\sqrt{2} $$x
$$ \Rightarrow $$ x = $$\frac{A}{\sqrt{2} } $$ = A cos $$\omega $$t
$$ \Rightarrow $$ cos $$\omega $$t = $$\frac{1}{\sqrt{2} } $$
$$ \Rightarrow $$ $$\omega $$t = $$\frac{\pi }{4} $$
$$ \Rightarrow $$ $$\frac{2\pi }{T} \cdot t=\frac{\pi }{4} $$
$$ \Rightarrow $$ t = $$\frac{T}{8} $$
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