Question 224: Oscillations

Question

A particle is performing S.H.M. starting from extreme position. Graphical representation shows that between displacement and acceleration, there is a phase difference of

Accepted Answer

Step 1: Recall the SHM relations
For a particle in SHM,
$ x(t) = A \cos(\omega t + \phi) $
where:
$x(t)$ = displacement
$A$ = amplitude
$\omega$ = angular frequency
$\phi$ = phase constant
The acceleration is related to displacement by:
$ a(t) = \ddot{x}(t) = -\omega^2 x(t) $
Step 2: Relationship between $x(t)$ and $a(t)$
From the expression:
$ a(t) = -\omega^2 x(t) $
This shows that
acceleration is directly proportional to displacement, but opposite in direction.
Step 3: Phase difference
Both $a(t)$ and $x(t)$ vary sinusoidally with the
same frequency
.
However, $a(t)$ has a
negative sign
compared to displacement.
This corresponds to a
phase difference of $\pi$ radians (180ÃÂ°).
Ã¢ÂÂ
Final Answer:
$ 	ext{Phase difference} = \pi \,	ext{rad} $
Correct option: A. $\pi \, 	ext{rad}$

Answer

$\pi \mathrm{rad}$

Answer

$\frac{\pi}{2} \mathrm{rad}$

Answer

$\frac{\pi}{4} \mathrm{rad}$

Answer

0 rad

Oscillations

A particle is performing S.H.M. starting from extreme position. Graphical representation shows that between displacement and acceleration, there is a phase difference of

Choose the correct answer: