Question 96: Oscillations

Question

The 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 = $${{3T} \over 4}$$
(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 = $${{T} \over 2}$$

Accepted Answer

(A) F = ma and a = $$-\omega^{2} x$$
At $$\frac{3T}{4} $$ displacement zero (x = 0), so a = 0
$$ 	herefore $$ 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
$$ 	herefore $$ V = A$$\omega $$ = V
max
(D) KE = PE
$$\frac{1}{2} k\left( A^{2}-x^{2}ight)  =\frac{1}{2} kx^{2}$$
$$ \Rightarrow $$ A
2
Ã¢ÂÂ x
2
= x
2
$$ \Rightarrow $$ A
2
= 2x
2
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 = $$\frac{T}{8} $$

Answer

(B), (C) and (D)

Answer

(A), (B) and (C)

Answer

(A) and (D)

Answer

(A), (B) and (D)

Oscillations

Choose the correct answer: