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 = ${{T} \over 2}$
![](https://imagex.cdn.examgoal.net/jgy2xukexsjhud/c0763f8e-d12e-4a01-a5ff-8661e682a5eb/56d38450-f3ee-11ea-8293-3b6cc591631c/file-jgy2xukexsjhue-320w.png)

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: